# NCERT Solutions Class 7 Maths Chapter 12

## NCERT Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions

The only way to be good at Mathematics is to solve more and more problems. In order to help students with their Mathematics preparation, Extramarks offers detailed and accurate NCERT Solutions for Class 7 Mathematics Chapter 12. Students can use these solutions as a guide to refer to whenever they get stuck with any of the NCERT textbook problems.

The solutions provided by Extramarks are prepared by subject matter experts who give special care and attention to providing accurate and step-by-step answers that will help students with their studies. The best way to use these solutions is for students to first attempt all the textbook problems on their own and then use these solutions to cross-check their answers and approach.

### Access NCERT Solutions for Class 7 Mathematics Chapter 12 – Algebraic Expressions and Identities

 Chapter 12 – Algebraic Expressions Exercise Exercise 12.1 Questions & Solutions Exercise 12.2 Questions & Solutions Exercise 12.3 Questions & Solutions Exercise 12.4 Questions & Solutions

## NCERT Solutions for Class 7 Mathematics Algebraic Expressions –

Chapter 12 of the Class 7 Mathematics NCERT textbook introduces students to Algebraic Expressions. Students learn about the important concepts of constants, coefficients, variables, factors, like, and unlike terms. The chapter also explains the methods to solve the algebraic expressions that involve one or two variables.

The chapter covers the following topics:

• How are Expressions Formed?
• Terms of an Expression
• Like and Unlike Terms
• Monomials, Binomials, Trinomials, and Polynomials
• Addition and Subtraction of Algebraic Expressions
• Finding The Value of an Expression
• Using Algebraic Expressions – Formulas and Rules

### Fundamental Concepts of Algebra

There are two essential terms in Algebra:

1. Constant
2. Variable

Constant: Constant is a quantity that has a fixed numerical value.

Example: 2, -5, 0.6, ¾, 8 etc.

Variable: Variable is the symbol that may have various numerical values based on given conditions. For this reason, these symbols are known as variables or lateral.

Example: Consider the formula to calculate the perimeter of a square

Formula = 4 × S (Sides)

In the formula, 4 is the constant point. But the length of the sides can be different at times.

When the side length is square is 4 cm, the perimeter will be (4×4) cm = 16 cm

If the side length of the square is 10 cm,

The perimeter will be (10×4) cm = 40 cm.

Here, we can say that the side and the perimeter are variables.

### Facts

Here we present a few facts associated with algebraic expressions:

• Constant has a fixed value, but the value of the variables differs in given situations.
• Any term of an algebraic expression has factors. The factors are variables. The numbers are the constant factors.
• You can create algebraic expressions using fundamental operations on variables and constants.
• A term has two factors. Any factor of a term is known as the coefficient of the remaining part of the term.
• When an algebraic expression has only one term, it is a monomial.
• When an algebraic expression has more than two terms, it is called a polynomial.
• When a polynomial has two terms, it is called binomial. If it has three, it is called trinomial.
• The terms with the same algebraic factors are called like terms, and those with different algebraic factors are known as, unlike terms.
• The students learn writing rules and formulas in concise forms using algebraic expressions.

### How Are Algebraic Expressions Formed?

An algebraic expression is a combination of constants and variables connected with symbols.

Constants are: 1, 2, 3, 4, -5, -8, -1

Variables are: x, y, z, a, b, c

In algebraic multiplication, the symbol is not necessary to use. One can write like am, which means a × m

On the other side, 2a means a × 2.

The method to form algebraic expressions is simple. Both constants and variables are used in the algebraic expressions.

Example:

1. 2x + 8
2. 12y – 6
3. 6 ×3y
4. 5bc + 7

Terms of an Expression

The students need to part the terms separately and then add them. The parts of an expression that are separated first and then added together are called terms.

1. 4x – 10

In this expression, 4x and 10 are two individual terms

1. 10a + 3

In this expression, 10a and 3 are two terms.

Factors of a Term

When two or more quantities are multiplied, each is known as a factor of the product. The quantities can be both numbers and literals.

1. 4xy

In this term, 4 is the numerical factor.

x and y are literal factors.

1. b3mn

In this term, b, m, and n are the literal factors.

3 is the numerical factor.

Coefficients

In a term, a numerical factor is called the coefficient of the remaining factors.

Example: 10abc

10 is the coefficient of abc.

Like and Unlike Terms

When the terms have the same algebraic factors, they are called like terms, and when they do not have the same algebraic factors, they are called ‘unlike’ terms.

Example: In the expression 3xy + 2x – 3y + 5xy

The terms 3xy and 5xy are like terms since they have the same algebraic factors xy.

The terms 2x and -3y are unlike terms since they have different algebraic factors.

Monomials, Binomials, Trinomials, and Polynomials

Monomials: When an expression has only one term.

Monomials Examples: 2x, 4pqr

Binomials:  When an expression has two terms.

Binomials Examples: a – 3, a + b

Trinomials: When an expression has three terms.

Trinomial Examples: x+ y+ z, a- b+ c

Polynomials: Polynomial is an algebraic expression that has two or more terms. Binomial and trinomial are polynomials.

Polynomial Examples: x + y, 4a – 3b + 1c + 5d

## NCERT Solutions for Class 7 Mathematics

To master and score good marks in Mathematics, one needs to put in a lot of effort and practise a sufficient number of problems on a daily basis With NCERT Solutions of Class 7 Mathematics, Extramarks tries to provide students with a valuable resource that they can use to get better at the subject and eventually score well in their final exams. Students can use these solutions if they ever get stuck on a problem. Moreover, the fundamentals of the chapter can be understood by the students in a better way.

## NCERT Solutions for Class 7

Apart from Mathematics, Extramarks provides detailed and accurate solutions to textbook questions for all of the other subjects covered in Class 7. All the solutions are prepared by their respective subject experts which give students an idea of how to attempt a question in the board exam in the right manner. Students can download all these solutions from the Extramarks’ website or app.

Q.1

$\begin{array}{l}\mathrm{Get}\mathrm{the}\mathrm{algebraic}\mathrm{expressions}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{cases}\mathrm{using}\\ \mathrm{variables},\mathrm{constants}\mathrm{and}\mathrm{arithmetic}\mathrm{operations}.\\ \left(\mathrm{i}\right)\mathrm{Subtraction}\mathrm{of}\mathrm{z}\mathrm{from}\mathrm{y}.\\ \left(\mathrm{ii}\right)\mathrm{One}–\mathrm{half}\mathrm{of}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{numbers}\mathrm{x}\mathrm{and}\mathrm{y}.\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{number}\mathrm{z}\mathrm{multiplied}\mathrm{by}\mathrm{itself}.\\ \left(\mathrm{iv}\right)\mathrm{One}–\mathrm{fourth}\mathrm{of}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{numbers}\mathrm{p}\mathrm{and}\mathrm{q}.\\ \left(\mathrm{v}\right)\mathrm{Numbersxandyboth}\mathrm{squared}\mathrm{and}\mathrm{added}.\\ \left(\mathrm{vi}\right)\mathrm{Number}5\mathrm{added}\mathrm{to}\mathrm{three}\mathrm{times}\mathrm{the}\mathrm{product}\mathrm{of}\\ \mathrm{numbers}\mathrm{m}\mathrm{and}\mathrm{n}.\\ \left(\mathrm{vii}\right)\mathrm{Product}\mathrm{of}\mathrm{numbers}\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{subtracted}\mathrm{from}10.\\ \left(\mathrm{viii}\right)\mathrm{Sum}\mathrm{of}\mathrm{numbers}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{subtracted}\mathrm{from}\mathrm{their}\\ \mathrm{product}.\end{array}$

Ans

$\begin{array}{l}\text{(i) y}-\text{z}\\ \text{(ii)}\frac{1}{2}\left(x+y\right)\end{array}$ $\begin{array}{l}{\text{(iii) z}}^{2}\\ \left(\text{iv)}\frac{1}{4}\left(pq\right)\\ {\text{(v) x}}^{2}+{y}^{2}\\ \text{(vi) 5+3}\left(mn\right)\\ \text{(vii) 10}-\text{yz}\\ \text{(viii) ab}-\left(a+b\right)\end{array}$

Q.2

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Identify}\mathrm{the}\mathrm{terms}\mathrm{and}\mathrm{their}\mathrm{factors}\mathrm{in}\mathrm{the}\mathrm{following}\\ \mathrm{expressions}.\\ \mathrm{Show}\mathrm{the}\mathrm{terms}\mathrm{and}\mathrm{factors}\mathrm{by}\mathrm{tree}\mathrm{diagrams}.\\ \left(\mathrm{a}\right)\mathrm{x}–3\left(\mathrm{b}\right)1+\mathrm{x}+{\mathrm{x}}^{2}\left(\mathrm{c}\right)\mathrm{y}–{\mathrm{y}}^{3}\\ \left(\mathrm{d}\right)5{\mathrm{xy}}^{2}+7{\mathrm{x}}^{2}\mathrm{y}\left(\mathrm{e}\right)–\mathrm{ab}+2{\mathrm{b}}^{2}–3{\mathrm{a}}^{2}\\ \left(\mathrm{ii}\right)\mathrm{Identify}\mathrm{terms}\mathrm{and}\mathrm{factors}\mathrm{in}\mathrm{the}\mathrm{expressions}\mathrm{given}\\ \mathrm{below}:\\ \left(\mathrm{a}\right)–4\mathrm{x}+5\left(\mathrm{b}\right)–4\mathrm{x}+5\mathrm{y}\left(\mathrm{c}\right)5\mathrm{y}+3{\mathrm{y}}^{2}\\ \left(\mathrm{d}\right)\mathrm{xy}+2{\mathrm{x}}^{2}{\mathrm{y}}^{2}\left(\mathrm{e}\right)\mathrm{pq}+\mathrm{q}\left(\mathrm{f}\right)1.2\mathrm{ab}–2.4\mathrm{b}+3.6\mathrm{a}\\ \left(\mathrm{g}\right)\frac{3}{4}\mathrm{x}+\frac{1}{4}\left(\mathrm{h}\right)0.1{\mathrm{p}}^{2}+0.2{\mathrm{q}}^{2}\end{array}$

Ans

(i)

a.

b.

c.

d.

e.

$\begin{array}{l}\left(\text{ii}\right)\\ \begin{array}{cccc}\text{Row}& \text{Expression}& \text{Terms}& \text{Factors}\\ \left(\text{a}\right)& -4x+5& \begin{array}{l}-4x\\ \text{5}\end{array}& \begin{array}{l}-4,x\\ 5\end{array}\\ \left(\text{b}\right)& -4x+5y& \begin{array}{l}-4x\\ 5y\end{array}& \begin{array}{l}-4,x\\ 5,y\end{array}\\ \left(\text{c}\right)& 5y+3{y}^{2}& \begin{array}{l}5y\\ 3{y}^{2}\end{array}& \begin{array}{l}5,y\\ 3,y,y\end{array}\\ \left(\text{d}\right)& xy+2{x}^{2}{y}^{2}& \begin{array}{l}xy\\ 2{x}^{2}{y}^{2}\end{array}& \begin{array}{l}x,y\\ 2,x,x,y,y\end{array}\\ \left(\text{e}\right)& pq+q& \begin{array}{l}pq\\ q\end{array}& \begin{array}{l}p,q\\ q\end{array}\\ \left(\text{f}\right)& 1.2ab-2.4b+3.6a& \begin{array}{l}1.2ab\\ -2.4b\\ 3.6a\end{array}& \begin{array}{l}1.2,a,b\\ -2.4,b\\ 3.6,a\end{array}\\ \left(\text{g}\right)& \frac{3}{4}x+\frac{1}{4}& \begin{array}{l}\frac{3}{4}x\\ \frac{1}{4}\end{array}& \begin{array}{l}\frac{3}{4},x\\ \frac{1}{4}\end{array}\\ \left(\text{h}\right)& 0.1{p}^{2}+0.2{q}^{2}& \begin{array}{l}0.1{p}^{2}\\ 0.2{q}^{2}\end{array}& \begin{array}{l}0.1,p,p\\ 0.2,q,q\end{array}\end{array}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Identify}\mathrm{the}\mathrm{numerical}\mathrm{coefficients}\mathrm{of}\mathrm{terms}\\ \left(\mathrm{other}\mathrm{than}\mathrm{constants}\right)\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{expressions}:\\ \left(\mathrm{i}\right)5–3{\mathrm{t}}^{2}\left(\mathrm{ii}\right)1+\mathrm{t}+{\mathrm{t}}^{2}+{\mathrm{t}}^{3}\left(\mathrm{iii}\right)\mathrm{x}+2\mathrm{xy}+3\mathrm{y}\\ \left(\mathrm{iv}\right)100\mathrm{m}+1000\mathrm{n}\left(\mathrm{v}\right)–{\mathrm{p}}^{2}{\mathrm{q}}^{2}+7\mathrm{pq}\left(\mathrm{vi}\right)1.2\mathrm{a}+0.8\mathrm{b}\\ \left(\mathrm{vii}\right)3.14{\mathrm{r}}^{2}\left(\mathrm{viii}\right)2\left(\mathrm{l}+\mathrm{b}\right)\left(\mathrm{ix}\right)0.1\mathrm{y}+0.01{\mathrm{y}}^{2}\end{array}$

Ans

$\begin{array}{cccc}\text{Row}& \text{Expression}& \text{Terms}& \text{Coefficients}\\ \left(\text{i}\right)& 5-3{\text{t}}^{2}& -3{t}^{2}& -3\\ \left(\text{ii}\right)& 1+t+{t}^{2}+{t}^{3}& \begin{array}{l}t\\ {t}^{2}\\ {t}^{3}\end{array}& \begin{array}{l}1\\ 1\\ 1\end{array}\\ \left(\text{iii}\right)& x+2xy+3y& \begin{array}{l}x\\ 2xy\\ 3y\end{array}& \begin{array}{l}1\\ 2\\ 3\end{array}\\ \left(\text{iv}\right)& 100m+100n& \begin{array}{l}100m\\ 100n\end{array}& \begin{array}{l}100\\ 100\end{array}\\ \left(\text{v}\right)& -{p}^{2}{q}^{2}+7pq& \begin{array}{l}-{p}^{2}{q}^{2}\\ 7pq\end{array}& \begin{array}{l}-1\\ 7\end{array}\\ \left(\text{vi}\right)& 1.2a+0.8b& \begin{array}{l}1.2a\\ 0.8b\end{array}& \begin{array}{l}1.2\\ 0.8\end{array}\\ \left(\text{vii}\right)& 3.14{r}^{2}& 3.14{r}^{2}& 3.14\\ \left(\text{viii}\right)& 2\left(l+b\right)& \begin{array}{l}2l\\ 2b\end{array}& \begin{array}{l}2\\ 2\end{array}\\ \left(\text{ix}\right)& 0.1y+0.01{y}^{2}& \begin{array}{l}0.1y\\ 0.01{y}^{2}\end{array}& \begin{array}{l}0.1\\ 0.01\end{array}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Identify}\mathrm{the}\mathrm{numerical}\mathrm{coefficients}\mathrm{of}\mathrm{terms}\\ \left(\mathrm{other}\mathrm{than}\mathrm{constants}\right)\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{expressions}:\\ \left(\mathrm{i}\right)5–3{\mathrm{t}}^{2}\left(\mathrm{ii}\right)1+\mathrm{t}+{\mathrm{t}}^{2}+{\mathrm{t}}^{3}\left(\mathrm{iii}\right)\mathrm{x}+2\mathrm{xy}+3\mathrm{y}\\ \left(\mathrm{iv}\right)100\mathrm{m}+1000\mathrm{n}\left(\mathrm{v}\right)–{\mathrm{p}}^{2}{\mathrm{q}}^{2}+7\mathrm{pq}\left(\mathrm{vi}\right)1.2\mathrm{a}+0.8\mathrm{b}\\ \left(\mathrm{vii}\right)3.14{\mathrm{r}}^{2}\left(\mathrm{viii}\right)2\left(\mathrm{l}+\mathrm{b}\right)\left(\mathrm{ix}\right)0.1\mathrm{y}+0.01{\mathrm{y}}^{2}\end{array}$

Ans

$\begin{array}{cccc}\text{Row}& \text{Expression}& \text{Terms}& \text{Coefficients}\\ \left(\text{i}\right)& 5-3{\text{t}}^{2}& -3{t}^{2}& -3\\ \left(\text{ii}\right)& 1+t+{t}^{2}+{t}^{3}& \begin{array}{l}t\\ {t}^{2}\\ {t}^{3}\end{array}& \begin{array}{l}1\\ 1\\ 1\end{array}\\ \left(\text{iii}\right)& x+2xy+3y& \begin{array}{l}x\\ 2xy\\ 3y\end{array}& \begin{array}{l}1\\ 2\\ 3\end{array}\\ \left(\text{iv}\right)& 100m+100n& \begin{array}{l}100m\\ 100n\end{array}& \begin{array}{l}100\\ 100\end{array}\\ \left(\text{v}\right)& -{p}^{2}{q}^{2}+7pq& \begin{array}{l}-{p}^{2}{q}^{2}\\ 7pq\end{array}& \begin{array}{l}-1\\ 7\end{array}\\ \left(\text{vi}\right)& 1.2a+0.8b& \begin{array}{l}1.2a\\ 0.8b\end{array}& \begin{array}{l}1.2\\ 0.8\end{array}\\ \left(\text{vii}\right)& 3.14{r}^{2}& 3.14{r}^{2}& 3.14\\ \left(\text{viii}\right)& 2\left(l+b\right)& \begin{array}{l}2l\\ 2b\end{array}& \begin{array}{l}2\\ 2\end{array}\\ \left(\text{ix}\right)& 0.1y+0.01{y}^{2}& \begin{array}{l}0.1y\\ 0.01{y}^{2}\end{array}& \begin{array}{l}0.1\\ 0.01\end{array}\end{array}$

Q.5

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Identify}\mathrm{terms}\mathrm{which}\mathrm{contain}\mathrm{x}\mathrm{and}\mathrm{give}\mathrm{the}\\ \mathrm{coefficient}\mathrm{of}\mathrm{x}.\\ \left(\mathrm{i}\right){\mathrm{y}}^{2}\mathrm{x}+\mathrm{y}\left(\mathrm{ii}\right)13{\mathrm{y}}^{2}–8\mathrm{yx}\left(\mathrm{iii}\right)\mathrm{x}+\mathrm{y}+2\\ \left(\mathrm{iv}\right)5+\mathrm{z}+\mathrm{zx}\left(\mathrm{v}\right)1+\mathrm{x}+\mathrm{xy}\left(\mathrm{vi}\right)12{\mathrm{xy}}^{2}+25\\ \left(\mathrm{vii}\right)7\mathrm{x}+{\mathrm{xy}}^{2}\\ \left(\mathrm{b}\right)\mathrm{Identify}\mathrm{terms}\mathrm{which}\mathrm{contain}{\mathrm{y}}^{2}\mathrm{and}\mathrm{give}\mathrm{the}\\ \mathrm{coefficient}\mathrm{of}{\mathrm{y}}^{2}.\\ \left(\mathrm{i}\right)8–{\mathrm{xy}}^{2}\left(\mathrm{ii}\right)5{\mathrm{y}}^{2}+7\mathrm{x}\left(\mathrm{iii}\right)2{\mathrm{x}}^{2}\mathrm{y}–15{\mathrm{xy}}^{2}+7{\mathrm{y}}^{2}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\\ \begin{array}{cccc}\text{Row}& \text{Expression}& \text{Terms with x}& \text{Cofficient of x}\\ \left(\text{i}\right)& {y}^{2}x+y& {y}^{2}x& {y}^{2}\\ \left(\text{ii}\right)& 13{y}^{2}-8yx& -8y& -8\\ \left(\text{iii}\right)& x+y+2& x& 1\\ \left(\text{iv}\right)& 5+z+zx& zx& z\\ \left(\text{v}\right)& 1+x+xy& \begin{array}{l}x\\ xy\end{array}& \begin{array}{l}1\\ y\end{array}\\ \left(\text{vi}\right)& 12x{y}^{2}+25& 12x{y}^{2}& 12{y}^{2}\\ \left(\text{vii}\right)& 7+x{y}^{2}& x{y}^{2}& {y}^{2}\end{array}\end{array}$ $\begin{array}{l}\left(\text{b}\right)\\ \begin{array}{cccc}\text{Row}& \text{Expression}& {\text{Terms with y}}^{2}& {\text{Cofficient of y}}^{2}\\ \left(\text{i}\right)& 8-x{y}^{2}& -x{y}^{2}& -x\\ \left(\text{ii}\right)& 5{y}^{2}+7x& 5{y}^{2}& 5\\ \left(\text{iii}\right)& 2{x}^{2}y+7{y}^{2}-15x{y}^{2}& \begin{array}{l}7{y}^{2}\\ -15x{y}^{2}\end{array}& \begin{array}{l}7\\ -15x\end{array}\end{array}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Classify}\mathrm{into}\mathrm{monomials},\mathrm{binomials}\mathrm{and}\mathrm{trinomials}.\\ \left(\mathrm{i}\right)4\mathrm{y}–7\mathrm{z}\left(\mathrm{ii}\right){\mathrm{y}}^{2}\left(\mathrm{iii}\right)\mathrm{x}+\mathrm{y}–\mathrm{xy}\left(\mathrm{iv}\right)100\\ \left(\mathrm{v}\right)\mathrm{ab}–\mathrm{a}–\mathrm{b}\left(\mathrm{vi}\right)5–3\mathrm{t}\left(\mathrm{vii}\right)4{\mathrm{p}}^{2}\mathrm{q}–4{\mathrm{pq}}^{2}\left(\mathrm{viii}\right)7\mathrm{mn}\\ \left(\mathrm{ix}\right){\mathrm{z}}^{2}–3\mathrm{z}+8\left(\mathrm{x}\right){\mathrm{a}}^{2}+{\mathrm{b}}^{2}\left(\mathrm{xi}\right){\mathrm{z}}^{2}+\mathrm{z}\left(\mathrm{xii}\right)1+\mathrm{x}+{\mathrm{x}}^{2}\end{array}$

Ans

$\begin{array}{l}\text{We know that the monomials, binomials and trinomials}\\ \text{have 1, 2, and 3 respectively unlike terms in it}\text{.}\\ \text{(i) 4y}-\text{7z}\\ \text{It is binomial}\text{.}\\ \text{(ii)}{y}^{2}\\ \text{It is monomial}\text{.}\\ \text{(iii) x}+\text{y}-\text{xy}\\ \text{It is trinomial}\text{.}\end{array}$ $\begin{array}{l}\text{(iv)}100\\ \text{It is monomial}\text{.}\\ \text{(v) ab}-\text{a}-\text{b}\\ \text{It is trinomial}\text{.}\\ \text{(vi) 5}-\text{3t}\\ \text{It is binomial}\text{.}\\ {\text{(vii) 4p}}^{2}q-{\text{4pq}}^{2}\\ \text{It is binomial}\text{.}\\ \text{(viii)}7mn\\ \text{It is monomial}\text{.}\\ {\text{(ix) z}}^{2}-\text{3z+8}\\ \text{It is trinomial}\text{.}\\ {\text{(x) a}}^{2}-{\text{b}}^{2}\\ \text{It is binomial}\text{.}\\ {\text{(xi) z}}^{2}+\text{z}\\ \text{It is binomial}\text{.}\\ \text{(xii) 1}+x+{\text{x}}^{2}\\ \text{It is trinomial}\text{.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{State}\mathrm{whether}\mathrm{a}\mathrm{given}\mathrm{pair}\mathrm{of}\mathrm{terms}\mathrm{is}\mathrm{of}\mathrm{like}\mathrm{or}\mathrm{unlike}\\ \mathrm{terms}.\\ \left(\mathrm{i}\right)1,100\left(\mathrm{ii}\right)–7\mathrm{x},52\mathrm{x}\left(\mathrm{iii}\right)–29\mathrm{x},–29\mathrm{y}\\ \left(\mathrm{iv}\right)14\mathrm{xy},42\mathrm{yx}\left(\mathrm{v}\right)4{\mathrm{m}}^{2}\mathrm{p},4{\mathrm{mp}}^{2}\left(\mathrm{vi}\right)12\mathrm{xz},12{\mathrm{x}}^{2}\mathrm{z}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{terms}\mathrm{which}\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{algebraic}\mathrm{factors}\mathrm{are}\mathrm{called}\\ \mathrm{like}\mathrm{terms}\mathrm{and}\mathrm{when}\mathrm{the}\mathrm{terms}\mathrm{have}\mathrm{different}\mathrm{algebraicf}\mathrm{actors},\\ \mathrm{they}\mathrm{are}\mathrm{called}\mathrm{unlike}\mathrm{terms}\\ \left(\mathrm{i}\right)1,100\\ \text{Like}\\ \left(\mathrm{ii}\right)–7\mathrm{x},52\mathrm{x}\\ \text{Like}\\ \left(\mathrm{iii}\right)–29\mathrm{x},–29\mathrm{y}\\ \text{UnLike}\\ \left(\mathrm{iv}\right)14\mathrm{xy},42\mathrm{yx}\\ \text{Like}\\ \left(\mathrm{v}\right)4{\mathrm{m}}^{2}\mathrm{p},4{\mathrm{mp}}^{2}\\ \text{UnLike}\\ \left(\mathrm{vi}\right)12\mathrm{xz},12{\mathrm{x}}^{2}{\mathrm{z}}^{2}\\ \mathrm{Un}\text{Like}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Identify}\mathrm{like}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)–{\mathrm{xy}}^{2},–4{\mathrm{yx}}^{2},8{\mathrm{x}}^{2},2{\mathrm{xy}}^{2},7\mathrm{y},–11{\mathrm{x}}^{2},–100\mathrm{x},–11\mathrm{yx},\\ 20{\mathrm{x}}^{2}\mathrm{y},–6{\mathrm{x}}^{2},\mathrm{y},2\mathrm{xy},3\mathrm{x}\\ \left(\mathrm{b}\right)10\mathrm{pq},7\mathrm{p},8\mathrm{q},–{\mathrm{p}}^{2}{\mathrm{q}}^{2},–7\mathrm{qp},–100\mathrm{q},–23,12{\mathrm{q}}^{2}{\mathrm{p}}^{2},\\ –5{\mathrm{p}}^{2},41,2405\mathrm{p},78\mathrm{qp},13{\mathrm{p}}^{2}\mathrm{q},{\mathrm{qp}}^{2},701{\mathrm{p}}^{2}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{Like terms are:}\\ -{\mathrm{xy}}^{2},2{\mathrm{xy}}^{2}\\ -4{\mathrm{yx}}^{2},20{\mathrm{x}}^{2}\mathrm{y}\\ 8{\mathrm{x}}^{2},-11{\mathrm{x}}^{2},-6{\mathrm{x}}^{2}\\ 7\mathrm{y},\mathrm{y}\\ -100\mathrm{x},3\mathrm{x}\\ -11\mathrm{xy},2\mathrm{xy}\\ \left(\mathrm{b}\right)\mathrm{Liketermsare}:\\ 10\mathrm{pq},-7\mathrm{pq},78\mathrm{qp}\\ 7\mathrm{p},2405\mathrm{p}\\ 8\mathrm{q},-100\mathrm{q}\\ -{\mathrm{p}}^{2}{\mathrm{q}}^{2},12{\mathrm{q}}^{2}{\mathrm{p}}^{2},\\ -23,41\\ -5{\mathrm{p}}^{2},701{\mathrm{p}}^{2}\\ 13{\mathrm{p}}^{2}\mathrm{q},{\mathrm{qp}}^{2}\end{array}$

Q.9

$\begin{array}{l}\mathrm{Simplify}\mathrm{combining}\mathrm{like}\mathrm{terms}:\\ \left(\mathrm{i}\right)21\mathrm{b}–32+7\mathrm{b}–20\mathrm{b}\\ \left(\mathrm{ii}\right)–{\mathrm{z}}^{2}+13{\mathrm{z}}^{2}–5\mathrm{z}+7{\mathrm{z}}^{3}–15\mathrm{z}\\ \left(\mathrm{iii}\right)\mathrm{p}–\left(\mathrm{p}–\mathrm{q}\right)–\mathrm{q}–\left(\mathrm{q}–\mathrm{p}\right)\\ \left(\mathrm{iv}\right)3\mathrm{a}–2\mathrm{b}–\mathrm{ab}–\left(\mathrm{a}–\mathrm{b}+\mathrm{ab}\right)+3\mathrm{ab}+\mathrm{b}–\mathrm{a}\\ \left(\mathrm{v}\right)5{\mathrm{x}}^{2}\mathrm{y}–5{\mathrm{x}}^{2}+3{\mathrm{yx}}^{2}–3{\mathrm{y}}^{2}+{\mathrm{x}}^{2}–{\mathrm{y}}^{2}+8{\mathrm{xy}}^{2}–3{\mathrm{y}}^{2}\\ \left(\mathrm{vi}\right)\left(3{\mathrm{y}}^{2}+5\mathrm{y}–4\right)–\left(8\mathrm{y}–{\mathrm{y}}^{2}–4\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)21\mathrm{b}–32+7\mathrm{b}–20\mathrm{b}\\ =21\mathrm{b}+7\mathrm{b}-20\mathrm{b}-32\\ =8\mathrm{b}-32\\ \left(\mathrm{ii}\right)-{\mathrm{z}}^{2}+13{\mathrm{z}}^{2}–5\mathrm{z}+7{\mathrm{z}}^{3}–15\mathrm{z}\\ =7{\mathrm{z}}^{3}+12{\mathrm{z}}^{2}-20\mathrm{z}\\ \left(\mathrm{iii}\right)\mathrm{p}-\left(\mathrm{p}-\mathrm{q}\right)-\mathrm{q}-\left(\mathrm{q}–\mathrm{p}\right)\\ =\overline{)\mathrm{p}}-\overline{)\mathrm{p}}+\overline{)\mathrm{q}}-\overline{)\mathrm{q}}-\mathrm{q}+\mathrm{p}\\ =\mathrm{p}-\mathrm{q}\end{array}$ $\begin{array}{l}\left(\mathrm{iv}\right)3\mathrm{a}-2\mathrm{b}-\mathrm{ab}-\left(\mathrm{a}-\mathrm{b}+\mathrm{ab}\right)+3\mathrm{ab}+\mathrm{b}-\mathrm{a}\\ =3\mathrm{a}-2\mathrm{b}-\mathrm{ab}-\mathrm{a}+\mathrm{b}-\mathrm{ab}+3\mathrm{ab}+\mathrm{b}-\mathrm{a}\\ =3\mathrm{a}-\mathrm{a}-\mathrm{a}-2\mathrm{b}+\mathrm{b}+\mathrm{b}+3\mathrm{ab}-\mathrm{ab}-\mathrm{ab}\\ =\mathrm{a}+\mathrm{ab}\\ \left(\mathrm{v}\right)5{\mathrm{x}}^{2}\mathrm{y}-5{\mathrm{x}}^{2}+3{\mathrm{yx}}^{2}-3{\mathrm{y}}^{2}+{\mathrm{x}}^{2}-{\mathrm{y}}^{2}+8{\mathrm{xy}}^{2}-3{\mathrm{y}}^{2}\\ =5{\mathrm{x}}^{2}\mathrm{y}+3{\mathrm{x}}^{2}\mathrm{y}-5{\mathrm{x}}^{2}+{\mathrm{x}}^{2}-3{\mathrm{y}}^{2}-{\mathrm{y}}^{2}-3{\mathrm{y}}^{2}+8{\mathrm{xy}}^{2}\\ =8{\mathrm{x}}^{2}\mathrm{y}-4{\mathrm{x}}^{2}-7{\mathrm{y}}^{2}+8{\mathrm{xy}}^{2}\\ \left(\mathrm{vi}\right)\left(3{\mathrm{y}}^{2}+5\mathrm{y}–4\right)–\left(8\mathrm{y}–{\mathrm{y}}^{2}–4\right)\\ =3{\mathrm{y}}^{2}+{\mathrm{y}}^{2}+5\mathrm{y}-8\mathrm{y}-4+4\\ =4{\mathrm{y}}^{2}-3\mathrm{y}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Add}:\\ \left(\mathrm{i}\right)3\mathrm{mn},–5\mathrm{mn},8\mathrm{mn},–4\mathrm{mn}\\ \left(\mathrm{ii}\right)\mathrm{t}–8\mathrm{tz},3\mathrm{tz}–\mathrm{z},\mathrm{z}–\mathrm{t}\\ \left(\mathrm{iii}\right)–7\mathrm{mn}+5,12\mathrm{mn}+2,9\mathrm{mn}–8,–2\mathrm{mn}–3\\ \left(\mathrm{iv}\right)\mathrm{a}+\mathrm{b}–3,\mathrm{b}–\mathrm{a}+3,\mathrm{a}–\mathrm{b}+3\\ \left(\mathrm{v}\right)14\mathrm{x}+10\mathrm{y}–12\mathrm{xy}–13,18–7\mathrm{x}–10\mathrm{y}+8\mathrm{xy},4\mathrm{xy}\\ \left(\mathrm{vi}\right)5\mathrm{m}–7\mathrm{n},3\mathrm{n}–4\mathrm{m}+2,2\mathrm{m}–3\mathrm{mn}–5\\ \left(\mathrm{vii}\right)4{\mathrm{x}}^{2}\mathrm{y},–3{\mathrm{xy}}^{2},–5{\mathrm{xy}}^{2},5{\mathrm{x}}^{2}\mathrm{y}\end{array}$ $\begin{array}{l}\left(\mathrm{viii}\right)3{\mathrm{p}}^{2}{\mathrm{q}}^{2}–4\mathrm{pq}+5,–10{\mathrm{p}}^{2}{\mathrm{q}}^{2},15+9\mathrm{pq}+7{\mathrm{p}}^{2}{\mathrm{q}}^{2}\\ \left(\mathrm{ix}\right)\mathrm{ab}–4\mathrm{a},4\mathrm{b}–\mathrm{ab},4\mathrm{a}–4\mathrm{b}\\ \left(\mathrm{x}\right){\mathrm{x}}^{2}–{\mathrm{y}}^{2}–1,{\mathrm{y}}^{2}–1–{\mathrm{x}}^{2},1–{\mathrm{x}}^{2}–{\mathrm{y}}^{2}\end{array}$

Ans

$\begin{array}{l}\left(i\right)\\ 3mn+\left(-5mn\right)+8mn+\left(-4mn\right)\\ =11mn-9mn\\ =2mn\\ \left(ii\right)\\ \left(t-8tz\right)+\left(3tz-z\right)+\left(z-t\right)\\ =\overline{)t}-8tz+3tz-\overline{)z}+\overline{)z}-\overline{)t}\\ =-5tz\\ \left(iii\right)\\ \left(-7mn+5\right)+\left(12mn+2\right)+\left(9mn-8\right)+\left(-2mn-3\right)\\ +-7mn+5+12mn+2+9mn-8-2mn-3\\ =12mn-4\\ \left(iv\right)\\ \left(a+b-3\right)+\left(b-a+3\right)+\left(a-b+3\right)\\ =\overline{)a}+\overline{)b}-\overline{)3}+b-\overline{)a}+\overline{)3}+a-\overline{)b}+3\\ =a+b+3\\ \left(vi\right)\\ \left(5m-7n\right)+\left(3n-4m+2\right)+\left(2m-3mn-5\right)\end{array}$ $\begin{array}{l}=5m-7n+3n-4m+2+2m-3mn-5\\ =3m-4n-3mn-3\\ \left(vii\right)\\ 4{x}^{2}y-3x{y}^{2}-5x{y}^{2}+5{x}^{2}y\\ =4{x}^{2}y+5{x}^{2}y-3x{y}^{2}-5x{y}^{2}\\ =9{x}^{2}y-8x{y}^{2}\\ \left(viii\right)\\ \left(3{p}^{2}{q}^{2}-4pq+5\right)+\left(-10{p}^{2}{q}^{2}\right)+\left(15+pq+7{p}^{2}{q}^{2}\right)\\ =3{p}^{2}{q}^{2}-10{p}^{2}{q}^{2}+7{p}^{2}{q}^{2}-4pq+9pq+5+15\\ =5pq+20\\ \left(ix\right)\\ \left(ab–4a\right)+\left(4b–ab\right)+\left(4a–4b\right)\\ =ab-ab-4a+4a+4b-4b\\ =0\\ \left(x\right)\\ \left({x}^{2}-{y}^{2}-1\right)+\left({y}^{2}-1-{x}^{2}\right)+\left(1-{x}^{2}-{y}^{2}\right)\\ =\overline{){x}^{2}}-\overline{){y}^{2}}-\overline{)1}+\overline{){y}^{2}}-\overline{)1}-\overline{){x}^{2}}+1-{x}^{2}-{y}^{2}\\ =-{x}^{2}-{y}^{2}-1\end{array}$

Q.11

$\begin{array}{l}\mathrm{Subtract}:\\ \left(\mathrm{i}\right)–5{\mathrm{y}}^{2}{\mathrm{fromy}}^{2}\\ \left(\mathrm{ii}\right)6\mathrm{xyfrom}–12\mathrm{xy}\\ \left(\mathrm{iii}\right)\left(\mathrm{a}–\mathrm{b}\right)\mathrm{from}\left(\mathrm{a}+\mathrm{b}\right)\\ \left(\mathrm{iv}\right)\mathrm{a}\left(\mathrm{b}–5\right)\mathrm{fromb}\left(5–\mathrm{a}\right)\\ \left(\mathrm{v}\right)–{\mathrm{m}}^{2}+5\mathrm{mnfrom}4{\mathrm{m}}^{2}–3\mathrm{mn}+8\\ \left(\mathrm{vi}\right)–{\mathrm{x}}^{2}+10\mathrm{x}–5\mathrm{from}5\mathrm{x}–10\\ \left(\mathrm{vii}\right)5{\mathrm{a}}^{2}–7\mathrm{ab}+5{\mathrm{b}}^{2}\mathrm{from}3\mathrm{ab}–2{\mathrm{a}}^{2}–2{\mathrm{b}}^{2}\\ \left(\mathrm{viii}\right)4\mathrm{pq}–5{\mathrm{q}}^{2}–3{\mathrm{p}}^{2}\mathrm{from}5{\mathrm{p}}^{2}+3{\mathrm{q}}^{2}–\mathrm{pq}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right){\mathrm{y}}^{2}-\left(-5{\mathrm{y}}^{2}\right)\\ =6{\mathrm{y}}^{2}\\ \left(\mathrm{ii}\right)-12\mathrm{xy}-\left(6\mathrm{xy}\right)\\ =-18\mathrm{xy}\\ \left(\mathrm{iii}\right)\left(\mathrm{a}+\mathrm{b}\right)-\left(\mathrm{a}-\mathrm{b}\right)\\ =\mathrm{a}+\mathrm{b}-\mathrm{a}+\mathrm{b}\\ =2\mathrm{b}\\ \left(\mathrm{iv}\right)\mathrm{b}\left(5-\mathrm{a}\right)-\mathrm{a}\left(\mathrm{b}-5\right)\\ =5\mathrm{b}-\mathrm{ab}-\mathrm{ab}+5\mathrm{a}\\ =5\mathrm{a}+5\mathrm{b}-2\mathrm{ab}\end{array}$ $\begin{array}{l}\left(\mathrm{v}\right)\left(4{\mathrm{m}}^{2}-3\mathrm{mn}+8\right)-\left(-{\mathrm{m}}^{2}+5\mathrm{mn}\right)\\ =4{\mathrm{m}}^{2}-3\mathrm{mn}+8+{\mathrm{m}}^{2}-5\mathrm{mn}\\ =5{\mathrm{m}}^{2}-8\mathrm{mn}+8\\ \left(\mathrm{vi}\right)\left(5\mathrm{x}-10\right)-\left(-{\mathrm{x}}^{2}+10\mathrm{x}-5\right)\\ =5\mathrm{x}-10+{\mathrm{x}}^{2}-10\mathrm{x}+5\\ ={\mathrm{x}}^{2}-5\mathrm{x}-5\\ \left(\mathrm{vii}\right)\left(3\mathrm{ab}-2{\mathrm{a}}^{2}-2{\mathrm{b}}^{2}\right)-\left(5{\mathrm{a}}^{2}-7\mathrm{ab}+5{\mathrm{b}}^{2}\right)\\ =3\mathrm{ab}-2{\mathrm{a}}^{2}-2{\mathrm{b}}^{2}-5{\mathrm{a}}^{2}+7\mathrm{ab}-5{\mathrm{b}}^{2}\\ =10\mathrm{ab}-7{\mathrm{a}}^{2}-7{\mathrm{b}}^{2}\\ \left(\mathrm{viii}\right)\left(5{\mathrm{p}}^{2}+3{\mathrm{q}}^{2}-\mathrm{pq}\right)-\left(4\mathrm{pq}-5{\mathrm{q}}^{2}-3{\mathrm{p}}^{2}\right)\\ =5{\mathrm{p}}^{2}+3{\mathrm{q}}^{2}-\mathrm{pq}-4\mathrm{pq}-5{\mathrm{q}}^{2}-3{\mathrm{p}}^{2}\\ =8{\mathrm{p}}^{2}+8{\mathrm{q}}^{2}-5\mathrm{pq}\end{array}$

Q.12

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{What}\mathrm{should}\mathrm{be}\mathrm{added}\mathrm{to}{\mathrm{x}}^{2}+\mathrm{xy}+{\mathrm{y}}^{2}\mathrm{to}\mathrm{obtain}2{\mathrm{x}}^{2}+3\mathrm{xy}?\\ \left(\mathrm{b}\right)\mathrm{What}\mathrm{should}\mathrm{be}\mathrm{subtracted}\mathrm{from}2\mathrm{a}+8\mathrm{b}+10\mathrm{to}\mathrm{get}–3\mathrm{a}+7\mathrm{b}+16?\end{array}$

Ans

$\begin{array}{l}\text{(a) Let a be the required sum}\text{.}\\ \text{Then, we have}\\ \text{a+}\left({\text{x}}^{\text{2}}{\text{+xy+y}}^{\text{2}}\right)=\left({\text{2x}}^{\text{2}}+\text{3xy}\right)\\ \text{a}={\text{2x}}^{2}+3\text{xy}-\left({\text{x}}^{2}+\text{xy}+{\text{y}}^{2}\right)\\ ={\text{2x}}^{2}+3\text{xy}-{\text{x}}^{2}-\text{xy}-{\text{y}}^{2}\\ ={\text{x}}^{2}-{\text{y}}^{2}+2\text{xy}\\ \text{(b) Let x be the required sum}\text{.}\\ \text{Then, we have}\\ \text{2a+8b+10}-\text{p}=-\text{3a+7b+16}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{p}=\text{2a+8b+10}-\text{(}-\text{3a+7b+16)}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2a+8b+10+3a}-\text{7b}-\text{16}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{5a+b}-\text{6}\end{array}$

Q.13 What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain –x2 – y2 + 6xy + 20?

Ans

$\begin{array}{l}Letpbetherequiredterm.\\ \left(3{x}^{2}-4{y}^{2}+5xy+20\right)-p=-{x}^{2}-{y}^{2}+6xy+20\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}p\text{\hspace{0.17em}}=\left(3{x}^{2}-4{y}^{2}+5xy+20\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}-\left(-{x}^{2}-{y}^{2}+6xy+20\right)\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3{x}^{2}-4{y}^{2}+5xy+20+{x}^{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}+{y}^{2}-6xy-20\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3{x}^{2}+{x}^{2}-4{y}^{2}+{y}^{2}+5xy\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}-6xy+20-20\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4{x}^{2}-3{y}^{2}-xy\end{array}$

Q.14

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{From}\mathrm{the}\mathrm{sum}\mathrm{of}3\mathrm{x}–\mathrm{y}+11\mathrm{and}–\mathrm{y}–11,\mathrm{subtract}3\mathrm{x}–\mathrm{y}–11.\\ \left(\mathrm{b}\right)\mathrm{From}\mathrm{the}\mathrm{sum}\mathrm{of}4+3\mathrm{x}\mathrm{and}5–4\mathrm{x}+2{\mathrm{x}}^{2},\mathrm{subtract}\mathrm{the}\mathrm{sum}\mathrm{of}3{\mathrm{x}}^{2}–5\mathrm{x}\mathrm{and}–{\mathrm{x}}^{2}+2\mathrm{x}+5.\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\left(3\mathrm{x}-\mathrm{}\mathrm{y}\mathrm{}+11\right)+\left(-\mathrm{}\mathrm{y}\mathrm{}-11\right)=3\mathrm{x}-\mathrm{y}\mathrm{}+11-\mathrm{y}\mathrm{}-11\\ =3\mathrm{x}\mathrm{}-\mathrm{y}\mathrm{}-\mathrm{}\mathrm{y}\mathrm{}+11-11\\ =3\mathrm{x}-2\mathrm{y}\\ \left(3\mathrm{x}\mathrm{}-2\mathrm{y}\right)-\left(3\mathrm{x}-\mathrm{}\mathrm{y}\mathrm{}-11\right)\\ =3\mathrm{x}\mathrm{}-2\mathrm{y}-3\mathrm{x}\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}+11\\ =3\mathrm{x}\mathrm{}-3\mathrm{x}\mathrm{}-2\mathrm{y}\mathrm{}+\mathrm{}\mathrm{y}\mathrm{}+11\\ =-\mathrm{}\mathrm{y}\mathrm{}+11\\ \left(\mathrm{b}\right)\left(4+3\mathrm{x}\right)+\left(5-4\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\right)\\ =4+3\mathrm{x}\mathrm{}+5-4\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\\ =3\mathrm{x}\mathrm{}-4\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\mathrm{}+4+5\\ =-\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\mathrm{}+9\\ \left(3{\mathrm{x}}^{2}-5\mathrm{x}\right)+\left(-{\mathrm{x}}^{2}\mathrm{}+2\mathrm{x}\mathrm{}+5\right)\\ =3{\mathrm{x}}^{2}-5\mathrm{x}\mathrm{}-{\mathrm{x}}^{2}\mathrm{}+2\mathrm{x}\mathrm{}+5\\ =3{\mathrm{x}}^{2}-\mathrm{}{\mathrm{x}}^{2\mathrm{}}-5\mathrm{x}\mathrm{}+2\mathrm{x}\mathrm{}+5\\ =2{\mathrm{x}}^{2}-3\mathrm{x}\mathrm{}+5\\ \left(-\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\mathrm{}+9\right)-\left(2{\mathrm{x}}^{2}-3\mathrm{x}\mathrm{}+5\right)\\ =-\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}\mathrm{}+9-2{\mathrm{x}}^{2}\mathrm{}+3\mathrm{x}-5\\ =-\mathrm{}\mathrm{x}\mathrm{}+3\mathrm{x}\mathrm{}+2{\mathrm{x}}^{2}-2{\mathrm{x}}^{2\mathrm{}}+9-5\\ =2\mathrm{x}\mathrm{}+4\end{array}$

Q.15

$\begin{array}{l}\mathrm{If}\mathrm{m}=2,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}:\\ \left(\mathrm{i}\right)\mathrm{m}–2\left(\mathrm{ii}\right)3\mathrm{m}–5\left(\mathrm{iii}\right)9–5\mathrm{m}\\ \left(\mathrm{iv}\right)3{\mathrm{m}}^{2}–2\mathrm{m}–7\left(\mathrm{v}\right)\frac{5\mathrm{m}}{2}–4\end{array}$

Ans

$\begin{array}{l}\mathrm{Form}=2\\ \left(\mathrm{i}\right)\mathrm{m}-2\\ =2-2\\ =0\\ \left(\mathrm{ii}\right)3\mathrm{m}-5\\ =3\left(2\right)-5\\ =6-5=1\\ \left(\mathrm{iii}\right)9-5\mathrm{m}\\ =9–5\left(2\right)\\ =9-10=-1\\ \left(\mathrm{iv}\right)3{\mathrm{m}}^{2}-2\mathrm{m}-7\\ =3{\left(2\right)}^{2}-2\left(2\right)-7\\ =12-4-7=1\end{array}$ $\begin{array}{l}\left(\mathrm{v}\right)\frac{5\mathrm{m}}{2}-4\\ =\frac{5\left(2\right)}{2}-4\\ =\frac{10}{2}-4\\ =5-4=1\end{array}$

Q.16

$\begin{array}{l}\mathrm{If}\mathrm{p}=–2,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}:\\ \left(\mathrm{i}\right)4\mathrm{p}+7\left(\mathrm{ii}\right)–3{\mathrm{p}}^{2}+4\mathrm{p}+7\left(\mathrm{iii}\right)–2{\mathrm{p}}^{3}–3{\mathrm{p}}^{2}+4\mathrm{p}+7\end{array}$

Ans

$\begin{array}{l}\mathrm{For}\mathrm{p}=-2,\\ \left(\mathrm{i}\right)4\mathrm{p}+7\\ =4\left(-2\right)+7\\ =-8+7=-1\\ \left(\mathrm{ii}\right)-3{\mathrm{p}}^{2}+4\mathrm{p}+7\\ =-3{\left(-2\right)}^{2}+4\left(-2\right)+7\\ =-3\left(4\right)-8+7\\ =-12-1=-13\end{array}$ $\begin{array}{l}\left(\mathrm{iii}\right)-2{\mathrm{p}}^{3}-3{\mathrm{p}}^{2}+4\mathrm{p}+7\\ =-2{\left(-2\right)}^{3}-3{\left(-2\right)}^{2}+4\left(-2\right)+7\\ =-2\left(-8\right)-3\left(4\right)-8+7\\ =16-12-1\\ =3\end{array}$

Q.17

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{expressions},\mathrm{when}\mathrm{x}=–1:\\ \left(\mathrm{i}\right)2\mathrm{x}–7\left(\mathrm{ii}\right)–\mathrm{x}+2\left(\mathrm{iii}\right){\mathrm{x}}^{2}+2\mathrm{x}+1\\ \left(\mathrm{iv}\right)2{\mathrm{x}}^{2}–\mathrm{x}–2\end{array}$

Ans

$\begin{array}{l}\mathrm{For}\mathrm{x}=-1:\\ \left(\mathrm{i}\right)2\mathrm{x}-7\\ =2\left(-1\right)-7\\ =-2-7=-9\\ \left(\mathrm{ii}\right)-\mathrm{x}+2\\ =-\left(-1\right)+2\\ =1+2=3\\ \left(\mathrm{iii}\right){\mathrm{x}}^{2}+2\mathrm{x}+1\\ ={\left(-1\right)}^{2}+2\left(-1\right)+1\\ =1-2+1=0\end{array}$ $\begin{array}{l}\left(\mathrm{iv}\right)2{\mathrm{x}}^{2}-\mathrm{x}-2\\ =2{\left(-1\right)}^{2}-\left(-1\right)-2\\ =2+1-2=1\end{array}$

Q.18

$\begin{array}{l}\mathrm{If}\mathrm{a}=2,\mathrm{b}=–2,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}:\\ \left(\mathrm{i}\right){\mathrm{a}}^{2}+{\mathrm{b}}^{2}\left(\mathrm{ii}\right){\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{b}}^{2}\left(\mathrm{iii}\right){\mathrm{a}}^{2}–{\mathrm{b}}^{2}\end{array}$

Ans

$\begin{array}{l}Fora=2,b=-2,\\ \left(i\right){a}^{2}+{b}^{2}\\ ={\left(2\right)}^{2}+{\left(-2\right)}^{2}\\ =4+4=8\\ \left(ii\right){a}^{2}+ab+{b}^{2}\\ ={\left(2\right)}^{2}+2\left(-2\right)+{\left(-2\right)}^{2}\\ =4-4+4=4\\ \left(iii\right){a}^{2}-{b}^{2}\\ ={\left(2\right)}^{2}-{\left(-2\right)}^{2}\\ =4-4=0\end{array}$

Q.19

$\begin{array}{l}\mathrm{When}\mathrm{a}=0,\mathrm{b}=–1,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{expressions}:\\ \left(\mathrm{i}\right)2\mathrm{a}+2\mathrm{b}\left(\mathrm{ii}\right)2{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+1\\ \left(\mathrm{iii}\right)2{\mathrm{a}}^{2}\mathrm{b}+2{\mathrm{ab}}^{2}+\mathrm{ab}\left(\mathrm{iv}\right){\mathrm{a}}^{2}+\mathrm{ab}+2\end{array}$

Ans

$\begin{array}{l}Fora=0,b=-1,\\ \left(i\right)2a+2b\\ =2\left(0\right)+2\left(-1\right)\\ =0-2=-2\\ \left(ii\right)2{a}^{2}+{b}^{2}+1\\ =2{\left(0\right)}^{2}+{\left(-1\right)}^{2}+1\\ =0+1+1=2\\ \left(iii\right)2{a}^{2}b+2a{b}^{2}+ab\\ =2{\left(0\right)}^{2}\left(-1\right)+2\left(0\right){\left(-1\right)}^{2}+0\left(-1\right)\\ =0+0+0=0\\ \left(iv\right){a}^{2}+ab+2\\ ={\left(0\right)}^{2}+\left(0\right)\left(-1\right)+2\\ =0+0+2=2\end{array}$

Q.20

$\begin{array}{l}\mathrm{Simplify}\mathrm{the}\mathrm{expressions}\mathrm{and}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{if}\mathrm{x}\mathrm{is}\mathrm{equal}\mathrm{to}2\\ \left(\mathrm{i}\right)\mathrm{x}+7+4\left(\mathrm{x}–5\right)\left(\mathrm{ii}\right)3\left(\mathrm{x}+2\right)+5\mathrm{x}–7\\ \left(\mathrm{iii}\right)6\mathrm{x}+5\left(\mathrm{x}–2\right)\left(\mathrm{iv}\right)4\left(2\mathrm{x}–1\right)+3\mathrm{x}+11\end{array}$

Ans

$\begin{array}{l}\left(i\right)x+7+4\left(x-5\right)\\ =x+7+4x-20\\ =5x-13\\ Forx=2\\ 5x-13\\ =5\left(2\right)-13\\ =10-13=-3\\ \left(ii\right)3\left(x+2\right)+5x-7\\ 3x+6+5x-7\\ =8x-1\\ Forx=2\\ =8\left(2\right)-1\\ =16-1=15\\ \left(iii\right)6x+5\left(x-2\right)\\ =6x+5x-10\\ =11x-10\end{array}$ $\begin{array}{l}Forx=2\\ =11\left(2\right)-10\\ =22-10=12\\ \left(iv\right)4\left(2x-1\right)+3x+11\\ =8x-4+3x+11\\ =11x+7\\ Forx=2\\ =11\left(2\right)+7\\ =22+7=29\end{array}$

Q.21

$\begin{array}{l}\mathrm{Simplify}\mathrm{these}\mathrm{expressions}\mathrm{and}\mathrm{find}\mathrm{their}\mathrm{values}\mathrm{if}\mathrm{x}=3,\\ \mathrm{a}=–1,\mathrm{b}=–2.\\ \left(\mathrm{i}\right)3\mathrm{x}–5–\mathrm{x}+9\left(\mathrm{ii}\right)2–8\mathrm{x}+4\mathrm{x}+4\\ \left(\mathrm{iii}\right)3\mathrm{a}+5–8\mathrm{a}+1\left(\mathrm{iv}\right)10–3\mathrm{b}–4–5\mathrm{b}\\ \left(\mathrm{v}\right)2\mathrm{a}–2\mathrm{b}–4–5+\mathrm{a}\end{array}$

Ans

$\begin{array}{l}\left(i\right)3x-5-x+9\\ =3x-x-5+9\\ =2x+4\\ Forx=3,a=-1,b=-2\end{array}$ $\begin{array}{l}2x+4\\ =2\left(3\right)+4\\ =6+4=10\\ \left(ii\right)2-8x+4x+4\\ =-4x+6\\ Forx=3,a=-1,b=-2\\ -4x+6\\ =-4\left(3\right)+6\\ =-12+6=-6\\ \left(iii\right)3a+5-8a+1\\ =-5a+6\\ Forx=3,a=-1,b=-2\\ -5a+6\\ =-5\left(-1\right)+6\\ =5+6=11\\ \left(iv\right)10-3b-4-5b\\ =10-4-3b-5b\\ =6-8b\\ Forx=3,a=-1,b=-2\\ 6-8b=6-8\left(-2\right)\\ =6+16=22\\ \left(v\right)2a-2b-4-5+a\\ =3a-2b-9\\ =3\left(-1\right)-2\left(-2\right)-9\\ =-3+4-9\\ =-8\end{array}$

Q.22

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{If}\mathrm{z}=10,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{z}}^{3}–3\left(\mathrm{z}–10\right).\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{p}=–10,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{p}}^{2}–2\mathrm{p}–100.\end{array}$

Ans

$\begin{array}{l}\left(i\right)Forz=10,\\ {z}^{3}–3\left(z–10\right)\\ ={\left(10\right)}^{3}-3\left(10-10\right)\\ =1000-3\left(0\right)=1000\\ \left(ii\right)Forp=-10,\\ {p}^{2}-2p-100\\ ={\left(-10\right)}^{2}-2\left(-10\right)-100\\ =100+20-100\\ =20\end{array}$

Q.23 What should be the value of a if the value of 2x2 +x – a equals to 5, when x = 0?

Ans

$\begin{array}{l}Given,2{x}^{2}+x-a=5,whenx=0\\ So,plugx=0toget\\ 2{\left(0\right)}^{2}+0-a=5\\ 0+0-a=5\\ \text{}\text{\hspace{0.17em}}-a=5\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=-5\end{array}$

Q.24

$\begin{array}{l}\mathrm{Simplify}\mathrm{the}\mathrm{expression}\mathrm{and}\mathrm{find}\mathrm{its}\mathrm{value}\mathrm{when}\mathrm{a}= 5\mathrm{and}\\ \mathrm{b}= –3.\\ 2\left({\mathrm{a}}^{\mathrm{2}}+\mathrm{ab}\right)+3–\mathrm{ab}\end{array}$

Ans

$\begin{array}{l}\text{Simplifying to get}\\ 2\left({a}^{2}+ab\right)+3-ab\\ =2{a}^{2}+2ab+3-ab\\ =2{a}^{2}+ab+3\end{array}$ $\begin{array}{l}whena=5andb=-3,\\ 2{\left(5\right)}^{2}+5\left(-3\right)+3\\ =50-15+3\\ =53-15\\ =38\end{array}$

Q.25

$\begin{array}{l}\mathrm{Observe}\mathrm{the}\mathrm{patterns}\mathrm{of}\mathrm{digits}\mathrm{made}\mathrm{from}\mathrm{line}\mathrm{segments}\\ \mathrm{of}\mathrm{equal}\mathrm{length}.\mathrm{You}\mathrm{will}\mathrm{findsuch}\mathrm{segmented}\mathrm{digits}\mathrm{on}\\ \mathrm{the}\mathrm{display}\mathrm{of}\mathrm{electronic}\mathrm{watches}\mathrm{or}\mathrm{calculators}.\end{array}$

If the number of digits of formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. Hoe many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.

Ans

a. It is given that the number of segments required to form n digits of the kind 6 is (5n + 1)

$\begin{array}{l}\text{Number of segments required to form 5 digits}\\ \text{=}\left(5×5+1\right)=26\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(5×10+1\right)=51\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(5×100+1\right)=501\end{array}$

b. It is given that the number of segments required to form n digits of the kind 4 is (3n + 1)

$\begin{array}{l}\text{Number of segments required to form 5 digits}\\ \text{=}\left(3×5+1\right)=16\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(3×10+1\right)=31\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(3×100+1\right)=301\end{array}$

c. It is given that the number of segments required to form n digits of the kind 8 is (5n + 2)

$\begin{array}{l}\text{Number of segments required to form 5 digits}\\ \text{=}\left(5×5+2\right)=27\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(5×10+2\right)=52\\ \text{Number of segments required to form 5 digits}\\ \text{=}\left(5×100+2\right)=502\end{array}$

Q.26 Use the given algebraic expression to complete the table of number patterns.

 S.No. Expression Terms 1st 2nd 3th 4th 5th 10th 100th (i) 1 3 5 7 9 – 19 – – – (ii) 2 5 8 11 – – – – – – (iii) 5 9 13 17 – – – – – – (iv) 27 34 41 48 – – – – – – (v) 2 5 10 17 – – – – – –

Ans

$\text{The given table can be completed as:}$

 S.No. Expression Terms 1st 2nd 3th 4th 5th … 10th 100th … (i) 1 3 5 7 9 – 19 – 199 – (ii) 2 5 8 11 – – – 32 302 – (iii) 5 9 13 17 – – – 41 401 – (iv) 27 34 41 48 – – – 90 720 – (v) 2 5 10 17 – – – 101 10,001 –