NCERT Solutions Class 10 Maths Chapter 3

Q.1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Ans. Let the present age of Aftab and his daughter be x years and y years respectively.
So, seven years ago,
Aftab’s age = (x–7) years and his daughter’s age = (y–7) years
According to the question,
x–7 = 7(y–7)
or x – 7y + 42 = 0
Three years hence,
Age of Aftab = (x + 3) years
Age of daughter = (y + 3) years
According to the question,
x + 3 = 3(y + 3)
or x – 3y – 6 = 0
Hence, the given information is represented algebraically by the two equations below.
x – 7y + 42 = 0
x – 3y – 6 = 0
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
x – 7y + 42 = 0
or x = 7y – 42

Also,
x – 3y – 6 = 0
or x = 3y + 6

The graphical representation is given below.

Q.2 The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.

Ans.

Let the price of a bat and a ball be ₹ x and ₹ y respectively.
According to the question,
3x + 6y = 3900 …(1)
or 3 (x + 2y) = 3900
or x + 2y = 3900/3 =1300 …(2)
Equation (1) represents the total price of 3 bats and 6 balls whereas equation (2) represents the total price of one bat and 2 balls.
To represent these equations graphically, we need at least two solutions for each equation. We write these solutions in table.
3x + 6y = 3900
or y = (3900 – 3x)/6

(i)
Also,
x + 2y =1300
or y = (1300 – x)/2

(ii)
The below given graphical representation shows that
graphs of both the equations coincide.

Q.3 The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.

Ans. Let the price of 1 kg of apple and 1 kg of grapes be ₹ x and ₹ y respectively.
According to the question,
2x + y = 160 …(1)
And
4x + 2y = 300
or 2x + y = 150 …(2)
Equations (1) and (2) represent the given situation algebraically.
To represent the given situation graphically, we need at least two solutions for each equation. We write these solutions in table.
2x + y = 160 …(1)
or y = 160 – 2x

(i)
Also,
2x + y = 150 …(2)
or y = 150 – 2x

(ii)
The graphical representation of the situation is given below. The two lines never intersect each other, i.e., the two lines are parallel.

Q.4 Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans.

(i) Let the number of girls and the number of boys be x and y respectively.
According to question,
x + y = 10
x – y = 4
For x + y = 10,
y = 10 – x

For x – y = 4,
y = x – 4

The graphs of equations are drawn below which shows that the two lines intersect at (7, 3).
Hence the number of boys and the number of girls are 7 and 3 respectively.

(ii)
Let the cost of 1 pencil and the cost of 1 pen be ₹ x and ₹ y respectively.
According to the question,
5x + 7y = 50
and
7x + 5y = 46
For 5x + 7y = 50,
y = (50 – 5x)/7

For 7x + 5y = 46,
y = (46 – 7x)/5

The graphs of equations are drawn below which shows that the two lines intersect at (3, 5).
Hence the cost of 1 pencil and the cost of 1 pen are ₹ 3 and ₹ 5 respectively.

Q.5 On comparing the ratios a1a2,b1b2, c1c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i)  5x – 4y + 8 = 0    7x + 6y – 9 = 0(ii) 9x + 3y + 12 = 0    18x + 6y + 24 = 0(iii) 6x – 3y + 10 = 0    2x – y + 9 = 0

Ans.

$\begin{array}{l}\text{(i) 5x-4y+8 = 0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7x+6y-9 = 0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{= 0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{=5,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{=-4,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{=8}\\ {\text{a}}_{\text{2}}{\text{=7,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{=6, \hspace{0.17em}c}}_{\text{2}}\text{=-9}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{5}}{\text{7}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{-4}}{\text{6}}\text{=}\frac{\text{-2}}{\text{3}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\ne \frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations intersect at a point.}\end{array}$ $\begin{array}{l}\text{(ii) 9x+3y+12=0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}18x+6y+24=0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{= 0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{=9,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{=3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{=12}\\ {\text{a}}_{\text{2}}{\text{=18,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{=6, c}}_{\text{2}}\text{=24}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{9}}{\text{18}}\text{=}\frac{\text{1}}{\text{2}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{3}}{\text{6}}\text{=}\frac{\text{1}}{\text{2}}\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\text{=}\frac{\text{12}}{\text{24}}\text{=}\frac{\text{1}}{\text{2}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations are coincident.}\end{array}$ $\begin{array}{l}\text{(iii) 6x-3y+10 = 0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x-y+9 = 0}\\ {\text{Comparing these equations with a}}_{\text{1}}{\text{x+b}}_{\text{1}}{\text{y+c}}_{\text{1}}\text{=0}\\ {\text{and a}}_{\text{2}}{\text{x+b}}_{\text{2}}{\text{y+c}}_{\text{2}}\text{= 0, we get}\\ {\text{a}}_{\text{1}}{\text{= 6,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{1}}{\text{= -3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}c}}_{\text{1}}\text{= 10}\\ {\text{a}}_{\text{2}}{\text{= 2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}_{\text{2}}{\text{= -1, c}}_{\text{2}}\text{= 9}\\ \text{Also,}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{\text{6}}{\text{2}}\text{= 3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\text{=}\frac{\text{-3}}{\text{-1}}\text{= 3,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\text{=}\frac{\text{10}}{\text{9}}\\ \text{and so}\\ \frac{{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{=}\frac{{\text{b}}_{\text{1}}}{{\text{\hspace{0.17em}b}}_{\text{2}}}\ne \frac{{\text{c}}_{\text{1}}}{{\text{\hspace{0.17em}c}}_{\text{2}}}\\ \text{Therefore, the given pairs of linear equations are parallel.}\end{array}$

Q.6

$\begin{array}{l}\text{On comparing the ratios}\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\text{,}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\text{,\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\text{, find out whether}\\ \text{the following pair of linear equations are consistent,}\\ \text{or inconsistent.}\\ \\ \text{(i)}3\mathrm{x}+2\mathrm{y}=5;2\mathrm{x}-3\mathrm{y}=7\\ \text{(ii)}2\mathrm{x}-3\mathrm{y}=8;4\mathrm{x}-6\mathrm{y}=9\\ \text{(iii)}\frac{3}{2}\mathrm{x}+\frac{5}{3}\mathrm{y}=7;9\mathrm{x}-10\mathrm{y}=14\\ \text{(iv) \hspace{0.17em}}5\mathrm{x}-3\mathrm{y}=11;-10\mathrm{x}+6\mathrm{y}=-22\\ \text{\hspace{0.17em}(v)}\frac{4}{3}\mathrm{x}+2\mathrm{y}=8;2\mathrm{x}+3\mathrm{y}=12\end{array}$

Ans.

$\begin{array}{l}\text{(i)}3\mathrm{x}+2\mathrm{y}=5;2\mathrm{x}-3\mathrm{y}=7\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=3,{\mathrm{b}}_1=2,{\mathrm{c}}_1=-5\\ {\mathrm{a}}_2=2,{\mathrm{b}}_2=-3,{\mathrm{c}}_2=-7\\ \text{Also,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{3}{2},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=-\frac{2}{3},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{5}{7}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \text{Therefore, the given pair of linear equations has a unique}\\ \text{solution and hence the two linear equations are consistent.}\end{array}$ \begin{array}{l}\text{(ii)}2x-3y=8;4x-6y=9\\ \text{Comparing these equations with}{a}_{1}x+b_{1}y+c_1=0\\ \text{and}{a}_{2}x+b_{2}y+c_2=0,\text{we get}\\ a_1=2,b_1=-3,c_1=-8\\ a_2=4,b_2=-6,c_2=-9\\ \text{Also,}\\ \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2},\frac{c_1}{c_2}=\frac{8}{9}\\ \text{and so}\\ \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\\ \text{Therefore, the given linear equations are parallel and}\\ \text{hence the two linear equations are inconsistent}\text{.}\end{array} $\begin{array}{l}\text{(iii)}\frac{3}{2}\mathrm{x}+\frac{5}{3}\mathrm{y}=7;9\mathrm{x}-10\mathrm{y}=14\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=\frac{3}{2},{\mathrm{b}}_1=\frac{5}{3},{\mathrm{c}}_1=-7\\ {\mathrm{a}}_2=9,{\mathrm{b}}_2=-10,{\mathrm{c}}_2=-14\\ \text{Also,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{\frac{3}{2}}{9}=\frac{1}{6},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{\frac{5}{3}}{-10}=-\frac{1}{6},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-7}{-14}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \text{Therefore, the given pair of linear equations has a unique}\\ \text{solution and hence the two linear equations are consistent.}\end{array}$ $\begin{array}{l}\text{(iv) \hspace{0.17em}}5\mathrm{x}-3\mathrm{y}=11;-10\mathrm{x}+6\mathrm{y}=-22\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=5,{\mathrm{b}}_1=-3,{\mathrm{c}}_1=-11\\ {\mathrm{a}}_2=-10,{\mathrm{b}}_2=6,{\mathrm{c}}_2=22\\ \text{Also,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{5}{-10}=-\frac{1}{2},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{-3}{6}=-\frac{1}{2},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-11}{22}=-\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they consistent.}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}(v)}\frac{4}{3}\mathrm{x}+2\mathrm{y}=8;2\mathrm{x}+3\mathrm{y}=12\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=\frac{4}{3},{\mathrm{b}}_1=2,{\mathrm{c}}_1=-8\\ {\mathrm{a}}_2=2,{\mathrm{b}}_2=3,{\mathrm{c}}_2=-12\\ \text{Also,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{2}{3},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-8}{-12}=\frac{2}{3}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they are consistent.}\end{array}$

Q.7

$\begin{array}{l}\text{Which of the following pairs of linear equations are\hspace{0.33em}consistent/inconsistent?}\\ \text{If consistent, obtain the\hspace{0.33em}solution graphically:}\\ \text{(i) \hspace{0.33em}\hspace{0.33em}x\hspace{0.33em}+\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}5,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2x\hspace{0.33em}+\hspace{0.33em}2y\hspace{0.33em}=\hspace{0.33em}10 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{(ii)\hspace{0.33em}\hspace{0.33em}x\hspace{0.33em}–\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}8,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x\hspace{0.33em}–\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}16}\\ \text{(iii)\hspace{0.33em}2x\hspace{0.33em}+\hspace{0.33em}y\hspace{0.33em}}–\text{6\hspace{0.33em}=\hspace{0.33em}0,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}4x\hspace{0.33em}–\hspace{0.33em}2y\hspace{0.33em}}–\text{\hspace{0.33em}4\hspace{0.33em}=\hspace{0.33em}0}\\ \text{(iv)\hspace{0.33em}2x\hspace{0.33em}–\hspace{0.33em}2y\hspace{0.33em}–\hspace{0.33em}2\hspace{0.33em}=\hspace{0.17em}0,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}4x\hspace{0.33em}–\hspace{0.33em}4y\hspace{0.33em}}–\text{\hspace{0.33em}5\hspace{0.33em}=\hspace{0.33em}0\hspace{0.17em}\hspace{0.17em}}\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}(i)}\mathrm{x}+\mathrm{y}=5,2\mathrm{x}+2\mathrm{y}=10\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=1,{\mathrm{b}}_1=1,{\mathrm{c}}_1=-5\\ {\mathrm{a}}_2=2,{\mathrm{b}}_2=2,{\mathrm{c}}_2=-10\\ \text{Now,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{2},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{1}{2},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-5}{-10}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Therefore, the given linear equations have infinitely many}\\ \text{solutions and hence they are consistent.}\\ \text{Graphs of the two equations coincide and hence each and}\\ \text{every point on this graph is a solution of these equations.}\end{array}$

$\begin{array}{l}\text{(ii)}\mathrm{x}-\mathrm{y}=8,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{x}-3\mathrm{y}=16\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=1,{\mathrm{b}}_1=-1,{\mathrm{c}}_1=-8\\ {\mathrm{a}}_2=3,{\mathrm{b}}_2=-3,{\mathrm{c}}_2=-16\\ \text{Now,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{3},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{1}{3},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-8}{-16}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Therefore, the given linear equations are parallel}\\ \text{and hence they are inconsistent.}\end{array}$ $\begin{array}{l}\text{(iii)}2\mathrm{x}+\mathrm{y}-6=0,4\mathrm{x}-2\mathrm{y}-4=0\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=2,{\mathrm{b}}_1=1,{\mathrm{c}}_1=-6\\ {\mathrm{a}}_2=4,{\mathrm{b}}_2=-2,{\mathrm{c}}_2=-4\\ \text{Now,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{4}=\frac{1}{2},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=-\frac{1}{2},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-6}{-4}=\frac{3}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \text{Therefore, the given linear equations are consistent.}\\ \text{Now,}\\ 2\mathrm{x}+\mathrm{y}-6=0\Rightarrow \mathrm{y}=6-2\mathrm{x}\\ \text{Some points which satisfy this equation are written}\\ \text{in the following table.}\\ \begin{array}{cccc}\mathrm{x}& 0& 2& 3\\ \mathrm{y}=6-2\mathrm{x}& 6& 2& 0\end{array}\\ \text{Again,}\\ 4\mathrm{x}-2\mathrm{y}-4=0\Rightarrow \mathrm{y}=2\mathrm{x}-2\\ \text{Some points which satisfy this equation are written}\\ \text{in the following table.}\\ \begin{array}{cccc}\mathrm{x}& 0& 2& 3\\ \mathrm{y}=2\mathrm{x}-2& -2& 2& 4\end{array}\\ \\ \text{We get the following graphs of the given equations and}\\ \text{find that they intersect at (2, 2). Hence, x = 2 and y = 2.}\end{array}$

$\begin{array}{l}\text{(iv) \hspace{0.17em}}2\mathrm{x}-2\mathrm{y}-2=0,4\mathrm{x}-4\mathrm{y}-5=0\\ \text{Comparing these equations with}{\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\\ \text{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\text{we get}\\ {\mathrm{a}}_1=2,{\mathrm{b}}_1=-2,{\mathrm{c}}_1=-2\\ {\mathrm{a}}_2=4,{\mathrm{b}}_2=-4,{\mathrm{c}}_2=-5\\ \text{Now,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{4}=\frac{1}{2},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{1}{3}=\frac{1}{2},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-2}{-5}=\frac{2}{5}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Therefore, the given linear equations are parallel}\\ \text{and hence they are inconsistent.}\end{array}$

Q.8 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans. Let the width of the garden be x and length be y.
According to the question, length is 4 m more than its width. Therefore,
y − x = 4 …(1)
Also, half the perimeter of a rectangular garden is 36 m.
Therefore,
x + y = 36 …(2)
Now,
y − x = 4 …(1)
or y = x + 4

Similarly,
x + y = 36 …(2)
or y = 36 – x

Graphs of the equations (1) and (2) are drawn below and from there we observe that they intersect at (16, 20).
Therefore, x = 16 and y = 20. Hence, width of the garden is 16 m and length of the garden is 20 m.

Q.9 Given the linear equation 2x+ 3y– 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Ans.

$\begin{array}{l}\text{(i) Intersecting lines:}\\ \text{A line which will intersect the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}3\mathrm{x}+2\mathrm{y}-8=0\text{as}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{3}\text{and}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{3}{2}\text{impllies that}\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}.\\ \text{(ii) P lines:}\\ \text{A line which is paralel to the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}4\mathrm{x}+6\mathrm{y}-7=0\text{as}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{4}=\frac{1}{2}\text{and}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{3}{6}=\frac{1}{2}\text{impllies that}\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}.\\ \left(\mathrm{iii}\right)\text{:}\\ \text{A line which is \hspace{0.17em}\hspace{0.17em}to the given line}2\mathrm{x}+3\mathrm{y}-8=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}is}4\mathrm{x}+6\mathrm{y}-16=0\text{because}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{1}{2}.\end{array}$

Q.10 Draw the graphs of the equations
x – y + 1 = 0 and 3x + 2y – 12 = 0.
Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Ans. We have,
x – y + 1 = 0
or y = x + 1
We have the following table.

Again,
3x + 2y – 12 = 0
or y = (12–3x)/2
and so we have the following table.

Graphs of the given equations are drawn below and from there we find that coordinates of the vertices of the triangle formed by these lines and the x-axis are (-1, 0); (4, 0) and (2, 3).

Q.11

$\begin{array}{l}\text{Solve the following pair of linear equations by the}\\ \text{substitution method.}\\ \text{(i)}\mathrm{x}+\mathrm{y}=14\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii)}\mathrm{s}-\mathrm{t}=3\\ \mathrm{x}-\mathrm{y}=4\frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iii)}3\mathrm{x}-\mathrm{y}=3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iv)}0.2\mathrm{x}+0.3\mathrm{y}=1.3\\ 9\mathrm{x}-3\mathrm{y}=90.4\mathrm{x}+0.5\mathrm{y}=2.3\\ \text{(v)}\sqrt{2}\mathrm{x}+\sqrt{3}\mathrm{y}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(vi)}\frac{3\mathrm{x}}{2}-\frac{5\mathrm{y}}{3}=-2\\ \sqrt{3}\mathrm{x}-\sqrt{8}\mathrm{y}=0\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{2}=\frac{13}{6}\end{array}$

Ans.

$\begin{array}{l}\text{(i) The given pair of linear equations are:}\\ \\ \mathrm{x}+\mathrm{y}=14...\left(1\right)\\ \mathrm{x}-\mathrm{y}=4...\left(2\right)\\ \text{We express x in terms of y from equation (1) to get}\\ \mathrm{x}=14-\mathrm{y}\\ \text{We substitute this value of x in equation (2) to get}\\ 14-\mathrm{y}-\mathrm{y}=4\\ \text{i.e.,}-2\mathrm{y}=4-14=-10\\ \text{i.e.,}\mathrm{y}=5\\ \text{Putting this value of y in equation (2), we get}\\ \mathrm{x}-5=4\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=9\\ \text{(ii) The given pair of linear equations are:}\\ \\ \mathrm{s}-\mathrm{t}=3...\left(1\right)\\ \frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6...\left(2\right)\\ \text{We express}\mathrm{s}\text{in terms of}\mathrm{t}\text{from equation (1) to get}\\ \mathrm{s}=\mathrm{t}+3\\ \text{We substitute this value of}\mathrm{s}\text{in equation (2) to get}\\ \frac{\mathrm{t}+3}{3}+\frac{\mathrm{t}}{2}=6\\ \text{i.e.,}\frac{5\mathrm{t}+6}{6}=6\\ \text{i.e.,}5\mathrm{t}+6=36\\ \text{i.e.,}\mathrm{t}=6\\ \text{Putting this value of t in equation (1), we get}\\ \mathrm{s}-6=3\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{s}=9\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}(iii) The given pair of linear equations are:}\\ \\ 3\mathrm{x}-\mathrm{y}=3...\left(1\right)\\ 9\mathrm{x}-3\mathrm{y}=9...\left(2\right)\\ \text{We express}\mathrm{y}\text{in terms of}\mathrm{x}\text{from equation (1) to get}\\ \mathrm{y}=3\mathrm{x}-3\\ \text{We substitute this value of}\mathrm{y}\text{in equation (2) to get}\\ 9\mathrm{x}-9\mathrm{x}+9=9\\ \text{i.e.,}9=9\\ \text{This statement is true for all values of x and so we can not}\\ \text{obtain a specific value of x. We observe that both the given}\\ \text{equations are the same as one is derived from another.}\\ \text{Therefore, given equations have infinitely many solutions.}\\ \text{\hspace{0.17em}(iv) The given pair of linear equations are:}\\ \\ 0.2\mathrm{x}+0.3\mathrm{y}=1.3...\left(1\right)\\ 0.4\mathrm{x}+0.5\mathrm{y}=2.3...\left(2\right)\\ \\ \text{We express x in terms of y from equation (1) to get}\\ \mathrm{x}=(1.3-0.3\mathrm{y})/(0.2)=(13-3\mathrm{y})/2\\ \text{We substitute this value of x in equation (2) to get}\\ \text{0.2}(13-3\mathrm{y})+0.5\mathrm{y}=2.3\\ \text{i.e.,}2.6-0.6\mathrm{y}+0.5\mathrm{y}=2.3\\ \text{i.e.,}-\text{0.1}\mathrm{y}=2.3-2.6=-0.3\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=3\\ \text{Putting this value of y in equation (2), we get}\\ 0.4\mathrm{x}+0.5\times 3=2.3\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=(2.3-1.5)/0.4=8/4=2\end{array}$ $\begin{array}{l}\left(\mathrm{v}\right)\mathrm{The}\mathrm{given}\mathrm{pair}\mathrm{of}\mathrm{linear}\mathrm{equations}\mathrm{are}:\\ \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sqrt{2}\mathrm{x}+\sqrt{3}\mathrm{y}=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(1\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sqrt{3}\mathrm{x}-\sqrt{8}\mathrm{y}=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(2\right)\\ \mathrm{We}\mathrm{express}\mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equation}(1)\mathrm{to}\mathrm{get}\\ \mathrm{x}=\frac{\sqrt{3}}{\sqrt{2}}\mathrm{y}\\ \mathrm{We}\mathrm{substitute}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}(2)\mathrm{to}\mathrm{get}\\ \sqrt{3}\frac{\sqrt{3}}{\sqrt{2}}\mathrm{y}-\sqrt{8}\mathrm{y}=0\\ \mathrm{i}.\mathrm{e}.,\frac{3-4}{\sqrt{2}}\mathrm{y}=0\\ \mathrm{i}.\mathrm{e}.,\mathrm{y}=0\\ \mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}(2),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sqrt{3}\mathrm{x}=0\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=0\\ \\ \left(\mathrm{vi}\right)\mathrm{The}\mathrm{given}\mathrm{pair}\mathrm{of}\mathrm{linear}\mathrm{equations}\mathrm{are}:\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{3\mathrm{x}}{2}-\frac{5\mathrm{y}}{3}=-2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ \mathrm{or}\hspace{0.17em}\hspace{0.17em}9\mathrm{x}-10\mathrm{y}\mathrm{\hspace{0.17em}}=-12\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(1\right)\mathrm{\hspace{0.17em}}\\ \mathrm{and}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{2}=\frac{13}{6}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\\ \mathrm{or}2\mathrm{x}+3\mathrm{y}=13\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}...\left(2\right)\\ \mathrm{We}\mathrm{express}\mathrm{x}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{y}\mathrm{from}\mathrm{equation}(1)\mathrm{to}\mathrm{get}\\ \mathrm{x}=(-12+10\mathrm{y})/9\\ \mathrm{We}\mathrm{substitute}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}(2)\mathrm{to}\mathrm{get}\\ 2(\frac{-12+10\mathrm{y}}{9})+3\mathrm{y}=13\\ \mathrm{i}.\mathrm{e}.,-24+20\mathrm{y}+27\mathrm{y}=117\\ \mathrm{i}.\mathrm{e}.,\; 47\mathrm{y}=117+24=141\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\frac{141}{47}=3\\ \mathrm{Putting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{in}\mathrm{equation}(2),\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\mathrm{x}+9=13\\ \mathrm{i}.\mathrm{e}.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=\frac{4}{2}=2\end{array}$

Q.12 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Ans.

$\begin{array}{l}\text{The given linear equations are:}\\ 2\mathrm{x}+3\mathrm{y}=11...\text{(1)}\\ 2\mathrm{x}-4\mathrm{y}=-24...\text{(2)}\\ \text{We express x in terms of y from equation (1) to get}\\ \mathrm{x}=\frac{11-3\mathrm{y}}{2}\\ \text{We substitute this value of x in equation (2) to get}\\ 11-3\mathrm{y}-4\mathrm{y}=-24\\ \text{i.e., 11}-7\mathrm{y}=-24\\ \text{i.e., 7}\mathrm{y}=24+11=35\\ \text{i.e., \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=5\\ \text{Putting this value of y in equation (1), we get}\\ 2\mathrm{x}+3\times 5=11\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}=11-15=-4\\ \text{i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=-2\\ \text{Now,}\\ \mathrm{y}=\mathrm{mx}+3\\ \text{or 5}=-2\mathrm{m}+3\\ \text{or}\mathrm{m}=-1\end{array}$

Q.13

$\begin{array}{l}\text{Form the pair of linear equations for the following}\\ \text{problems and find their solution by substitution method.}\\ \text{(i) The difference betrween two numbers is 26 and one}\\ \text{number is three times the other. Find them.}\\ \text{(ii) The larger of two supplementary angles exceeds}\\ \text{the smaller by 18 degrees. Find them.}\\ \text{(iii) The coach of a cricket team buys a 7 bats and 6 balls}\\ \text{for}₹\text{3800. Later she buys the 3 bats and 5 balls}\\ \text{for}₹\text{1750. Find the cost of each bat and each ball.}\\ \text{(iv) The taxi charges in a city consist of a fixed charge}\\ \text{together with the charge for the distance covered.}\\ \text{For\hspace{0.17em}\hspace{0.17em}a distance of 10 km, the charge paid is}₹\text{105 and}\\ \text{for a journey of 15 km, the charge paid is 155. What}\\ \text{are the\hspace{0.17em}\hspace{0.17em}fixed charges and the charge per km? How}\\ \text{much does\hspace{0.17em}\hspace{0.17em}a person have to pay for travelling a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}distance of 25 km?}\\ \text{(v) A fraction becomes}\frac{9}{11}\text{, if 2 is added to both the}\\ \text{numerator and the denominator. If, 3 is added to}\\ \text{both the numerator and the denominator it}\\ \text{becomes}\frac{5}{6}.\text{Find the fraction.}\\ \text{(vi) Five years hence, the age of Jacob will be three times}\\ \text{that of his son. Five years ago, Jacob’s age was seven}\\ \text{times that of his son. What are their present ages?}\end{array}$

Ans.

(i)
Let the larger number is y and the smaller number is x.
According to question,
y = 3x …(1)
and
y – x = 26 …(2)
We substitute the value of y from equation (1) in equation (2) and get
3x – x = 26
or 2x = 26
or x = 13
Putting this value of x in equation (1), we get
y = 39
Hence, the numbers are 13 and 39.

(ii)
Let the larger angle is y and the smaller angle is x.
According to question,
y – x = 18° …(1)
and
y + x = 180° …(2)
We write y in terms of x from (1) to get
y = 18° + x
Putting this value of y in equation (2), we get
18° + x + x = 180°
or 2x = 180° – 18° = 162°
or x = 81°
Putting this value of x in equation (1), we get
y = 18° + 81° = 99°
Hence, the two supplementary angles are 81° and 99°.

(iii)
Let the cost of one bat and one ball be ₹ x and ₹ y respectively.
According to question,
7x + 6y = ₹ 3800 …(1)
and
3x + 5y = ₹ 1750 …(2)
We write y in terms of x from (1) to get
y= (3800 – 7x)/6
Putting this value of y in equation (2), we get
3x + 5(3800 – 7x)/6 = 1750
or 18x – 35x = 1750 × 6 – 19000 = –8500
or x = 8500/17 = 500
Putting this value of x in equation (1), we get
7×500 + 6y = 3800
or y = (3800 – 3500)/6 = 50
Hence, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.

(iv)
Let the fixed charge be ₹ x and charge per km be ₹ y.
According to question,
x + 10y = ₹ 105 …(1)
and
x + 15y = ₹ 155 …(2)
We write x in terms of y from (1) to get
x = 105 – 10y
Putting this value of x in equation (2), we get
105 – 10y + 15y = 155
or 5y = 155 – 105 = 50
or y = 10
Putting this value of y in equation (1), we get
x + 10×10 = 105
or x = 105 – 100 = 5
Hence, the fixed charge is ₹ 5 and charge per km is ₹ 10.
Charge for 25 km = 5 + 25 ×10 = ₹ 255

$\begin{array}{l}\left(\text{v}\right)\\ \text{Let the fraction be}\frac{\mathrm{x}}{\mathrm{y}}.\\ \text{According to question,}\\ \frac{\mathrm{x}+2}{\mathrm{y}+2}=\frac{9}{11}\Rightarrow 11\mathrm{x}-9\mathrm{y}=-4...\left(1\right)\\ \text{and}\\ \frac{\mathrm{x}+3}{\mathrm{y}+3}=\frac{5}{6}\Rightarrow 6\mathrm{x}-5\mathrm{y}=-3...\left(2\right)\\ \\ \text{From equation (2), we get}\\ \mathrm{x}=\frac{5\mathrm{y}-3}{6}\\ \text{We put this value of}\mathrm{x}\text{in equation (1) and get}\\ 11\frac{5\mathrm{y}-3}{6}-9\mathrm{y}=-4\\ \Rightarrow 55\mathrm{y}-33-54\mathrm{y}=-24\\ \Rightarrow \mathrm{y}=-24+33=9\\ \text{Putting this value of y in equation (1), we get}\\ 11\mathrm{x}-9\times 9=-4\\ \Rightarrow 11\mathrm{x}=81-4=77\\ \Rightarrow \mathrm{x}=7\\ \text{Hence the fraction is}\frac{7}{9}.\end{array}$

(vi)
Let the age of Jacob be x and the age of his son be y.
According to question,
x + 5 = 3(y + 5)
or x – 3y = 10 …(1)
Also,
x – 5 = 7(y – 5)
or x – 7y = –30 …(2)
From equation (1), we find
x = 10 + 3y
We substitute this value of x in equation (2) and get
10 + 3y – 7y = –30
or –4y = –40
or y = 10
Putting this value of y in (1), we get
x – 3×10 = 10
or x = 40
Hence the present age of Jacob is 40 years and that of his son is 10 years.

Q.14 

$\begin{array}{l}\text{Solve the following pair of linear equations by the}\\ \text{elimination method and the substitution method:}\\ \text{(i) \hspace{0.33em}\hspace{0.33em}x\hspace{0.33em}+\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}5\hspace{0.33em}and\hspace{0.33em}2x\hspace{0.33em}–\hspace{0.33em}3y\hspace{0.33em}=\hspace{0.33em}4}\\ \text{(ii) \hspace{0.33em}3x\hspace{0.33em}+\hspace{0.33em}4y\hspace{0.33em}=\hspace{0.33em}10\hspace{0.33em}and\hspace{0.33em}2x\hspace{0.33em}–\hspace{0.33em}2y\hspace{0.33em}=\hspace{0.33em}2}\\ \text{(iii) 3x\hspace{0.33em}–\hspace{0.33em}5y\hspace{0.33em}–\hspace{0.33em}4\hspace{0.33em}=\hspace{0.33em}0\hspace{0.33em}and\hspace{0.33em}9x\hspace{0.33em}=\hspace{0.33em}2y\hspace{0.33em}+\hspace{0.33em}7}\\ \text{(iv)\hspace{0.33em}}\frac{\text{x}}{\text{2}}\text{\hspace{0.33em}+\hspace{0.33em}}\frac{\text{2y}}{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}–1\hspace{0.33em}and\hspace{0.33em}x\hspace{0.33em}–}\frac{\text{y}}{\text{3}}\text{\hspace{0.33em}=\hspace{0.33em}3}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{By elimination method}:\\ \text{Given that}\\ \mathrm{x}+\mathrm{y}=5...\text{(1)}\\ 2\mathrm{x}-3\mathrm{y}=4...\text{(2)}\\ \text{We multiply equation (1) by 3 and then add it to}\\ \text{equation (2) to eliminate the variable y and get}\\ \text{the value of x as follows.}\\ 3\mathrm{x}+3\mathrm{y}=15\\ \underset{¯}{2\mathrm{x}-3\mathrm{y}=4}\\ 5\mathrm{x}=19\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{19}{5}\\ \text{Using this value of x in equation (1), we get}\\ \frac{19}{5}+\mathrm{y}=5\\ \text{or}\mathrm{y}=5-\frac{19}{5}=\frac{6}{5}\\ \text{By substitution method}:\\ \text{From equation (1), we get}\\ \mathrm{y}=5-\mathrm{x}\\ \text{Substituting this in equation (2), we get}\\ 2\mathrm{x}-3(5-\mathrm{x})=4\\ \text{or}5\mathrm{x}=4+15=19\\ \text{or}\mathrm{x}=\frac{19}{5}\\ \text{Substituting this value of x in equation (1), we get}\\ \frac{19}{5}+\mathrm{y}=5\\ \text{or}\mathrm{y}=5-\frac{19}{5}=\frac{6}{5}\end{array}$ $\begin{array}{l}\text{(ii)}\\ \text{By elimination method}:\\ \text{Given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{x}+4\mathrm{y}=10...\text{(1)}\\ 2\mathrm{x}-2\mathrm{y}=2...\text{(2)}\\ \text{We multiply equation (2) by 2 and then add it to}\\ \text{equation (1) to eliminate the variable y and get}\\ \text{the value of x as follows.}\\ 3\mathrm{x}+4\mathrm{y}=10\\ \underset{¯}{4\mathrm{x}-4\mathrm{y}=4}\\ 7\mathrm{x}=14\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=2\\ \text{Using this value of x in equation (1), we get}\\ 6+4\mathrm{y}=10\\ \text{or}\mathrm{y}=\frac{10-6}{4}=1\\ \text{By substitution method}:\\ \text{From equation (2), we get}\\ \mathrm{x}=1+\mathrm{y}\\ \text{Substituting this in equation (1), we get}\\ 3(1+\mathrm{y})+4\mathrm{y}=10\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}7\mathrm{y}=10-3=7\\ \text{or}\mathrm{y}=1\\ \text{Substituting this value of y in equation (2), we get}\\ 2\mathrm{x}-2=2\\ \text{or}\mathrm{x}=\frac{2+2}{2}=2\end{array}$ $\begin{array}{l}\text{(iii)}\\ \text{By elimination method}:\\ \text{Given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}\mathrm{x}-5\mathrm{y}-4=0...\text{(1)}\\ 9\mathrm{x}=2\mathrm{y}+7...\text{(2)}\\ \text{We multiply equation (1) by 3 and then subtract}\\ \text{equation (2) from it to eliminate the variable x and get}\\ \text{the value of y as follows.}\\ 9\mathrm{x}-15\mathrm{y}-12=0\\ \underset{¯}{\begin{array}{l}9\mathrm{x}=2\mathrm{y}+7\\ ---\end{array}}\\ -15\mathrm{y}-12=-2\mathrm{y}-7\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-15\mathrm{y}+2\mathrm{y}=-7+12=5\\ \text{or}\mathrm{y}=-\frac{5}{13}\\ \text{Using this value of y in equation (2), we get}\\ 9\mathrm{x}=-\frac{5}{13}\times 2+7\\ \text{or}\mathrm{x}=\frac{-10}{117}+\frac{7}{9}=\frac{-10+91}{117}=\frac{81}{117}=\frac{9}{13}\\ \text{By substitution method}:\\ \text{From equation (2), we get}\\ \mathrm{x}=\frac{2\mathrm{y}+7}{9}\\ \text{Substituting this in equation (1), we get}\\ 3\times \frac{2\mathrm{y}+7}{9}-5\mathrm{y}-4=0\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{y}+7}{3}-5\mathrm{y}=4\\ \text{or}2\mathrm{y}+7-15\mathrm{y}=12\\ \text{or}-13\mathrm{y}=12-7\\ \text{or}\mathrm{y}=-\frac{5}{13}\\ \text{Substituting this value of y in equation (2), we get}\\ 9\mathrm{x}=-2\times \frac{5}{13}+7=\frac{-10+91}{13}\\ \text{or}\mathrm{x}=\frac{81}{13\times 9}=\frac{9}{13}\end{array}$ $\begin{array}{l}\text{(iv)}\\ \text{By elimination method}:\\ \text{Given that}\\ \frac{\mathrm{x}}{2}+\frac{2\mathrm{y}}{3}=-1...\text{(1)}\\ \mathrm{x}-\frac{\mathrm{y}}{3}=3...\text{(2)}\\ \text{We multiply equation (2) by 2 and then add it to}\\ \text{equation (1) to eliminate the variable y and get}\\ \text{the value of x as follows.}\\ \frac{\mathrm{x}}{2}+\frac{2\mathrm{y}}{3}=-1\\ \underset{¯}{2\mathrm{x}-\frac{2\mathrm{y}}{3}=6}\\ \frac{\mathrm{x}}{2}+2\mathrm{x}=5\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5\mathrm{x}}{2}=5\\ \text{or}\mathrm{x}=2\\ \text{Using this value of x in equation (2), we get}\\ 2-\frac{\mathrm{y}}{3}=3\\ \text{or}-\mathrm{y}=3(3-2)=3\\ \text{or}\mathrm{y}=-3\\ \text{By substitution method}:\\ \text{From equation (2), we get}\\ \mathrm{x}=3+\frac{\mathrm{y}}{3}\\ \text{Substituting this in equation (1), we get}\\ \frac{1}{2}(3+\frac{\mathrm{y}}{3})+\frac{2\mathrm{y}}{3}=-1\\ \text{or \hspace{0.17em}\hspace{0.17em}}\frac{9+\mathrm{y}}{6}+\frac{2\mathrm{y}}{3}=-1\\ \text{or}\frac{9+\mathrm{y}+4\mathrm{y}}{6}=-1\\ \text{or}9+\mathrm{y}+4\mathrm{y}=-6\\ \text{or}\mathrm{y}=\frac{-15}{5}=-3\\ \text{Substituting this value of y in equation (2), we get}\\ \mathrm{x}=3+\frac{-3}{3}=2\end{array}$

Q.15 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Let the fraction be}\frac{\mathrm{x}}{\mathrm{y}}\text{.}\\ \text{According to question,}\\ \frac{\mathrm{x}+1}{\mathrm{y}-1}=1\\ \Rightarrow \mathrm{x}+1=\mathrm{y}-1\\ \Rightarrow \mathrm{x}-\mathrm{y}=-2...\text{(1)}\\ \text{Also,}\\ \frac{\mathrm{x}}{\mathrm{y}+1}=\frac{1}{2}\\ \Rightarrow \mathrm{x}=\frac{\mathrm{y}+1}{2}\\ \Rightarrow \text{2}\mathrm{x}=\mathrm{y}+1\\ \Rightarrow \text{2}\mathrm{x}-\mathrm{y}=1...\left(2\right)\\ \text{Subtracting equation (2) from equation (1), we get}\\ \mathrm{x}=3\\ \text{Substituting this value in equation (1), we get}\\ \mathrm{y}=5\\ \text{Hence, the fraction is}\frac{3}{5}\text{.}\end{array}$ $\begin{array}{l}\text{(ii)}\\ \text{Let the present age of Nuri and Sonu be x years}\\ \text{and y years respectively}.\\ \text{So},\text{five years ago},\\ \text{Nuri}‘\text{s age}=(\text{x}–\text{5})\text{years}\\ \text{and}\\ \text{Sonu’s age}=(\text{y}–\text{5})\text{years}\\ \text{According to question,}\\ \mathrm{x}-5=3(\text{y}–\text{5})\\ \Rightarrow \mathrm{x}-5=3\mathrm{y}-15\\ \Rightarrow \mathrm{x}-3\mathrm{y}=-10...\left(1\right)\\ \text{Again,}\\ \text{Ten years hence},\\ \text{Nuri}‘\text{s age}=(\mathrm{x}+10)\text{years}\\ \text{and}\\ \text{Sonu’s age}=(\mathrm{y}+10)\text{years}\\ \text{According to the question},\\ \mathrm{x}+10=2(\mathrm{y}+10)\\ \text{or}\mathrm{x}-2\mathrm{y}=10...\left(2\right)\\ \text{Subtracting equation (2) from equation (1), we get}\\ \mathrm{y}=20\\ \text{Putting this value in (2), we get}\\ \mathrm{x}=50\\ \text{Hence, Nuri and Sonu are 50 years}\\ \text{and 20 years old respectively}.\end{array}$ $\begin{array}{l}\text{(iii)}\\ \text{Let the two digit number be}10\mathrm{x}+\mathrm{y}.\\ \text{According to question,}\\ \mathrm{x}+\mathrm{y}=9...\left(1\right)\\ \text{and}\\ 9(10\mathrm{x}+\mathrm{y})=2(10\mathrm{y}+\mathrm{x})\\ \Rightarrow \text{90x}+9\mathrm{y}=20\mathrm{y}+2\mathrm{x}\\ \Rightarrow 90\mathrm{x}-2\mathrm{x}+9\mathrm{y}-20\mathrm{y}=0\\ \Rightarrow 88\mathrm{x}-11\mathrm{y}=0\\ \Rightarrow 8\mathrm{x}-\mathrm{y}=0...\text{(2)}\\ \text{Adding equations (1) and (2), we get}\\ \text{x = 1}\\ \text{Putting this value of x in equation (2), we get}\\ \mathrm{y}=8\\ \text{Hence,}\\ \text{Number}=10\mathrm{x}+\mathrm{y}=10+8=18\end{array}$ $\begin{array}{l}\text{(iv)}\\ \text{Let number of notes of}₹50\text{and}₹100\mathrm{bexandyrespectively}.\\ \text{According to question,}\\ \mathrm{x}+\mathrm{y}=25...\left(1\right)\\ \text{and}\\ 50\mathrm{x}+100\mathrm{y}=2000\\ \Rightarrow \text{50\hspace{0.17em}}(\text{x}+2\mathrm{y})=2000\\ \Rightarrow \text{x}+2\mathrm{y}=40...\text{(2)}\\ \\ \text{Subtracting equation (1) from equation(2), we get}\\ \text{x}+2\mathrm{y}=40\\ \underset{¯}{\begin{array}{l}\mathrm{x}+\mathrm{y}=25\\ ---\end{array}}\\ \mathrm{y}=15\\ \text{Putting this value in equation (1), we get}\\ \mathrm{x}=10\\ \text{Hence,}\\ \text{Meena received 10 notes of}‘50\text{and 15 notes of}‘100.\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\left(\mathrm{v}\right)\\ \text{Let the fixed charge for first three day be}₹\mathrm{x}\text{and additional}\\ \text{charge for each day thereafter be}₹\mathrm{y}.\\ \text{According to question,}\\ \mathrm{x}+4\mathrm{y}=27...\left(1\right)\\ \text{and}\\ \mathrm{x}+2\mathrm{y}=21...\text{(2)}\\ \\ \text{Subtracting equation (2) from equation (1), we get}\\ \text{x}+4\mathrm{y}=27\\ \underset{¯}{\begin{array}{l}\mathrm{x}+2\mathrm{y}=21\\ ---\end{array}}\\ 2\mathrm{y}=6\\ \text{or}\mathrm{y}=3\\ \text{Putting this value in equation (1), we get}\\ \mathrm{x}=27-12=15\\ \text{Hence, fixed charge is}₹15\text{and additional charge for}\\ \text{each extra day is}₹3.\end{array}$

Q.16 

$\begin{array}{l}\text{Which of the following pairs of linear equations has\hspace{0.33em}unique solution,}\\ \text{no solution, or infinitely many\hspace{0.33em}solutions. In case there is a unique}\\ \text{solution,\hspace{0.33em}find it by using cross multiplication method.}\\ \text{(i) x\hspace{0.33em}}–\text{\hspace{0.33em}3y\hspace{0.33em}}–\text{3\hspace{0.33em}=\hspace{0.33em}0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(ii) 2x\hspace{0.33em}+\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}5}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x\hspace{0.33em}}–\text{\hspace{0.33em}9y\hspace{0.33em}}–\text{\hspace{0.33em}2\hspace{0.33em}=\hspace{0.33em}0\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x\hspace{0.33em}+\hspace{0.33em}2y\hspace{0.33em}=\hspace{0.33em}8}\\ \text{(iii) 3x\hspace{0.33em}}–\text{\hspace{0.33em}5y\hspace{0.33em}=\hspace{0.33em}20\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(iv)\hspace{0.17em}\hspace{0.17em}x\hspace{0.33em}}–\text{\hspace{0.33em}3y\hspace{0.33em}}–\text{\hspace{0.33em}7\hspace{0.33em}=\hspace{0.33em}0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}6x\hspace{0.33em}}–\text{\hspace{0.33em}10y\hspace{0.33em}=\hspace{0.33em}40\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x\hspace{0.33em}}–\text{\hspace{0.33em}3y\hspace{0.33em}}–\text{\hspace{0.33em}15\hspace{0.33em}=\hspace{0.33em}0}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\mathrm{x}-3\mathrm{y}-3=0\\ 3\mathrm{x}-9\mathrm{y}-2=0\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{3}\text{,}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{-3}{-9}=\frac{1}{3}\text{,}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-3}{-2}=\frac{3}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Thus, the given pair of linear equations have no solution.}\\ \text{(ii)}2\mathrm{x}+\mathrm{y}=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i.e., \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+\mathrm{y}-5=0\\ 3\mathrm{x}+2\mathrm{y}=8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}+2\mathrm{y}-8=0\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{3}\text{,}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{1}{2}\text{,}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{5}{8}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \text{Thus, the given pair of linear equations have unique solution.}\\ \text{By cross-multiplication method, we have}\\ \frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{a}}_1{\mathrm{c}}_2}=\frac{1}{{\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{b}}_1{\mathrm{a}}_2}\\ \Rightarrow \frac{\mathrm{x}}{-8-(-10)}=\frac{\mathrm{y}}{-15-(-16)}=\frac{1}{4-3}\\ \Rightarrow \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{1}=\frac{1}{1}\\ \Rightarrow \mathrm{x}=2\text{and}\mathrm{y}=1\end{array}$ $\begin{array}{l}\text{(iii)}3\mathrm{x}-5\mathrm{y}=20\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i.e., \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\mathrm{x}-5\mathrm{y}-20=0\\ 6\mathrm{x}-10\mathrm{y}=40\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}i.e.,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{x}-10\mathrm{y}-40=0\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{3}{6}=\frac{1}{2}\text{,}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{-5}{-10}=\frac{1}{2}\text{,}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-20}{-40}=\frac{1}{2}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{Thus, the given pair of linear equations have infinitely}\\ \text{many solutions.}\\ \text{(iv)}\mathrm{x}-3\mathrm{y}-7=0\\ 3\mathrm{x}-3\mathrm{y}-15=0\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{3}\text{,}\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{-3}{-3}=1\text{,}\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-7}{-15}=\frac{7}{15}\\ \text{and so}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\\ \text{Thus, the given pair of linear equations have unique solution.}\\ \text{By cross-multiplication method, we have}\\ \frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{a}}_1{\mathrm{c}}_2}=\frac{1}{{\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{b}}_1{\mathrm{a}}_2}\\ \Rightarrow \frac{\mathrm{x}}{45-21}=\frac{\mathrm{y}}{-21-(-15)}=\frac{1}{-3-(-9)}\\ \Rightarrow \frac{\mathrm{x}}{24}=\frac{\mathrm{y}}{-6}=\frac{1}{6}\\ \Rightarrow \mathrm{x}=4\text{and}\mathrm{y}=-1\end{array}$

Q.17

$\begin{array}{l}\text{(i) For which values of a and\hspace{0.33em}b does the following pair\hspace{0.33em}of linear equations have an}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}infinite number of\hspace{0.33em}solutions?}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}2x\hspace{0.33em}+\hspace{0.33em}3y\hspace{0.33em}=7}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(a\hspace{0.33em}–\hspace{0.33em}b)x\hspace{0.33em}+\hspace{0.33em}(a\hspace{0.33em}+\hspace{0.33em}b)y\hspace{0.33em}=\hspace{0.33em}3a\hspace{0.33em}+\hspace{0.33em}b\hspace{0.33em}}–\text{\hspace{0.33em}2}\\ \text{(ii) For which value of k will the following pair\hspace{0.33em}of linear equations have no solution?}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x\hspace{0.33em}+\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}1}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(2k\hspace{0.33em}–\hspace{0.33em}1)x\hspace{0.33em}+\hspace{0.33em}(k\hspace{0.33em}–\hspace{0.33em}1)\hspace{0.33em}y\hspace{0.33em}=\hspace{0.33em}2k\hspace{0.33em}+\hspace{0.33em}1}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Given pair of linear equations are:}\\ 2\mathrm{x}+3\mathrm{y}=7\\ (\mathrm{a}-\mathrm{b})\mathrm{x}+(\mathrm{a}+\mathrm{b})\mathrm{y}=3\mathrm{a}+\mathrm{b}-2\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{\mathrm{a}-\mathrm{b}},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{3}{\mathrm{a}+\mathrm{b}},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{7}{3\mathrm{a}+\mathrm{b}-2}\\ \text{For a pair of linear equations to have infinitely many solutions:}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{So, we need}\\ \frac{2}{\mathrm{a}-\mathrm{b}}=\frac{3}{\mathrm{a}+\mathrm{b}}=\frac{7}{3\mathrm{a}+\mathrm{b}-2}\\ \text{or}\frac{2}{\mathrm{a}-\mathrm{b}}=\frac{7}{3\mathrm{a}+\mathrm{b}-2}\text{\hspace{0.17em} and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2}{\mathrm{a}-\mathrm{b}}=\frac{3}{\mathrm{a}+\mathrm{b}}\\ \text{or}2(3\mathrm{a}+\mathrm{b}-2)=7(\mathrm{a}-\mathrm{b})\text{and \hspace{0.17em}\hspace{0.17em}}2(\mathrm{a}+\mathrm{b})=3(\mathrm{a}-\mathrm{b})\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{a}+2\mathrm{b}-4=7\mathrm{a}-7\mathrm{b}\text{and \hspace{0.17em}\hspace{0.17em}}2\mathrm{a}-3\mathrm{a}=-3\mathrm{b}-2\mathrm{b}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\mathrm{a}+9\mathrm{b}-4=0\text{and}\mathrm{a}=5\mathrm{b}\\ \text{Now, we put}\mathrm{a}=5\mathrm{b}\text{in \hspace{0.17em}}-\mathrm{a}+9\mathrm{b}-4=0.\\ \text{Thus,}\\ -5\mathrm{b}+9\mathrm{b}=4\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{b}=4\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}=1\\ \text{Also,}\\ \mathrm{a}=5\mathrm{b}=5\times 1=5\end{array}$ $\begin{array}{l}\text{(ii)}\\ \text{Given pair of linear equations are:}\\ 3\mathrm{x}+\mathrm{y}=1\\ (2\mathrm{k}-1)\mathrm{x}+(\mathrm{k}-1)\mathrm{y}=2\mathrm{k}+1\\ \text{Here,}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{3}{2\mathrm{k}-1},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{1}{\mathrm{k}-1},\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{1}{2\mathrm{k}+1}\\ \text{For a pair of linear equations to have no solution:}\\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}\\ \text{So, the given pair of linear equations have no solution when}\\ \frac{3}{2\mathrm{k}-1}=\frac{1}{\mathrm{k}-1}\\ \text{or}3\mathrm{k}-3=2\mathrm{k}-1\\ \text{or}3\mathrm{k}-2\mathrm{k}=-1+3\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=2\end{array}$

Q.18

$\begin{array}{l}\text{Solve the following pair of linear equations by the}\\ \text{substitution and cross-multiplication methods:}\\ 8\mathrm{x}+5\mathrm{y}=9\\ 3\mathrm{x}+2\mathrm{y}=4\end{array}$

Ans.

$\begin{array}{l}8\mathrm{x}+5\mathrm{y}=9...\left(1\right)\\ 3\mathrm{x}+2\mathrm{y}=4...\left(2\right)\\ \text{By substitution mwthod:}\\ \text{From equation (2), we get}\\ \mathrm{y}=\frac{4-3\mathrm{x}}{2}\\ \text{Substituting this value of}\mathrm{y}\text{in equation (1), we get}\\ 8\mathrm{x}+5\mathrm{y}=9\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\mathrm{x}+5\times \frac{4-3\mathrm{x}}{2}=9\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}16\mathrm{x}+20-15\mathrm{x}=18\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=18-20=-2\\ \text{Putting this value of x in equation (2), we get}\\ 3\times (-2)+2\mathrm{y}=4\\ \text{or}-6+2\mathrm{y}=4\\ \text{or 2y}=4+6=10\\ \text{or}\mathrm{y}=5\end{array}$ $\begin{array}{l}\text{By cross-multiplication method}:\\ 8\mathrm{x}+5\mathrm{y}=9\\ 3\mathrm{x}+2\mathrm{y}=4\\ \text{Or}\\ 8\mathrm{x}+5\mathrm{y}-9=0...\left(1\right)\\ 3\mathrm{x}+2\mathrm{y}-4=0...\left(2\right)\\ \\ {\text{Here, a}}_1=8,{\text{b}}_1{\text{= 5, c}}_1\text{=}-\text{9}\\ {\text{a}}_2=3,{\text{b}}_2{\text{= 2, c}}_2\text{=}-4\\ \text{By cross-multiplication method, we have}\\ \frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{a}}_1{\mathrm{c}}_2}=\frac{1}{{\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{b}}_1{\mathrm{a}}_2}\\ \Rightarrow \frac{\mathrm{x}}{-20+18}=\frac{\mathrm{y}}{-27+32}=\frac{1}{16-15}\\ \Rightarrow \frac{\mathrm{x}}{-2}=\frac{\mathrm{y}}{5}=\frac{1}{1}\\ \Rightarrow \mathrm{x}=-2\text{and}\mathrm{y}=5\end{array}$