NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions (EX 12.2) Exercise 12.2

Class 7 is an important academic year for students. The foundations of all core subjects are laid in this class. Therefore, it is very important that they pay attention to what is being taught in Class 7. The knowledge gained in Class 7 will be very useful for students throughout their lives. Understanding concepts is crucial for Class 7 students since it will enable them to think strategically and do better in their studies and exams. Even if achieving high marks or a high rank is not as important in Class 7, students should perform well because doing well in exams builds confidence and allows them to perform better in subsequent classes.

Numerous subjects are studied by Class 7 students. The core subjects in Class 7 are Mathematics, Social Studies, Science, Sanskrit/Hindi, and English. Students must give each subject in Class 7 the same amount of attention because they are all very important. The study resources provided by Extramarks can be used by students to prepare for every topic. Extramarks provides resources such as NCERT solutions, past years’ paper solutions, and revision notes. Students must use the provided resources in order to boost their Class 7 scores. The resources from Extramarks are really helpful for self-studying. Extramarks’ materials are extremely dependable. Each resource has been meticulously written by subject-matter specialists and is regularly proofread. Additionally, the resources are updated according to the most recent CBSE rules.

For Class 7 exams, students must prepare well. If students have a firm grasp of the subjects in Class 7, they can study effectively in higher classes. All exercise problems and the extracurricular activities listed in the NCERT textbooks for Class 7 must be completed by the students. This will enable students to gain useful subject knowledge. Students can use the materials on Extramarks if they face any difficulties.

Mathematics, the Science of structure, order, and relation, was born from the basic operations of counting, measuring, and describing the shapes of objects. It involves logical deduction and mathematical computations, and as it has developed, its subject matter has gotten more idealised and abstract. Since the 17th century, Mathematics has been an essential supplement to the Physical Sciences and technology.

Students often consider Mathematics to be challenging. Students must practise a significant number of questions in order to make Mathematics concise and simple to learn. Furthermore, consistency is necessary for a thorough understanding of Mathematics. To be fully prepared for Mathematics exams, students must answer questions with varying levels of complexity. They should begin the chapter by responding to simple questions before gradually moving on to those that call for them to employ their critical thinking and reasoning abilities.

Students should be prepared to answer a variety of questions because they will likely see more questions in assessments that require analytical thinking.They will be able to complete questions within the time limit with enough practice.Students are encouraged to read the NCERT book to study for Class 7 Mathematics.

The National Council of Educational Research and Training (NCERT) is a national level autonomous organisation formed by the government of India to provide advice to the central and state governments of India on policies and programmes to develop and improve school education in the country. NCERT was established in 1961 and is in charge of publishing instructional materials for Classes 1 through 12.

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NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions (EX 12.2) Exercise 12.2 

Chapter 12 of Class 7 Mathematics is on Algebraic Expressions. To ace in Mathematics, students must practice a lot of questions. Students can practise the questions of Class 7 Chapter 12 Exercise 12.2 from NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 are offered by Extramarks for the benefit of students. Students can download the  NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 from the website and mobile application of Extramarks. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 can be downloaded in PDF format for offline access.

With the help of the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 students will be able to solve the Class 7 Maths Chapter 12 Exercise 12.2 thoroughly. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 are written in a stepwise manner for the clarity of students. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 can help students to self study and enhance their Mathematics skills.

Access NCERT Solutions for Maths Class 7 Chapter 12- Algebraic Expressions

Students can access the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 from the website and mobile application of Extramarks. Students are advised to download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 in PDF format. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 will help students score well in the exams. Students must download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2  for effective preparation.

NCERT Solutions For Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2

According to the NCERT curriculum, Chapter 12 of Algebraic Expressions covers the ideas of Constants, Variables, Coefficients, Factors, and similar and unlike terms. Additionally, this chapter explains how to write and solve Algebraic Expressions with one or two variables. For solutions of Class 7th Maths Chapter 12 Exercise 12.2 students can refer to the  NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 are available on the website and mobile application of Extramarks.

The various topics and concepts covered in Class 12 Algebraic Expressions are:

In Arithmetic, quantities are represented by numbers with predetermined values, whereas in algebra, quantities are represented by symbols (often letters from the English alphabet) that can have several values. Additionally, these symbols’ values change depending on the circumstances. These are referred to as Variables.

In Algebra, there are two categories of terms:

  • Constant: A phrase with a set numerical value is referred to as a constant. Thus, the values 3, -2, 4/9, and 0.2 are constants.
  • Variable: A sign that can have different numerical values depending on the circumstances is referred to as a variable or a literal.

For instance, we are aware that the formula for a square’s perimeter is:

P = 4 x s

Since 4 is constant here,

P equals 4 x 6 = 24 for s = 6.

P equals 4 x 5 = 20 for s = 5.

This indicates that p and s’s values are not constant. They differ. S and P are variables in the aforementioned case.

Students can learn more about Constants and Variables and answer questions related to these concepts with the help of the  NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The  NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 will guide students to solve all the questions accurately.

In this chapter, students also learn about how Algebraic expressions are formed.The symbols +, —, x, and are used to connect variables, constants, or simply variables in an Algebraic Equation.

For instance:

The Variable y is multiplied by itself to produce the equation y3

y * y * y = y3

By multiplying the variable y by itself twice and the resultant product of y2 by the number 2, the expression 2y2 is created.

In order to produce the phrase (2y3 – 7), multiply the variable y by itself three times. Then, multiply the result of y3 by the constant 2, which gives 2y3. Next, take 7 out of the product to arrive at the final result (2y3 – 7).

The expression xy is obtained by multiplying the variable x with another variable y. Thus, x*y = xy.

To get expression 5xy + 7, students first need to obtain xy, next they multiply it by 5 we get 5xy. Then,7 is added to 5xy to get the expression 5xy + 7.

Terms of an Expression: Students also learn about Terms of Expressions in Chapter 12 Class 7 Mathematics. An Algebraic Expression is created by first separating its Components, and then adding them together. Terms are used to describe these expression components that are initially produced individually, and subsequently they are combined.

Factor: When two or more quantities (numbers and literals) are multiplied, each of them is referred to as a factor of the product. To understand what Factors are, and to solve questions related to it, students must refer to the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 will guide students to solve all the questions in the most precise manner. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 can be used to enhance student’s Mathematics scores in Class 7.

Coefficient: The Coefficient of the product of all the remaining factors is the name given to any one of a term’s factors.

Like and Unlike terms: Terms that share comparable literal aspects are referred to as like terms, whilst terms that do not share similar literal factors are referred to as unlike terms (or dissimilar terms).

Monomial: A Monomial is a phrase with just one term.

Binomial: A Binomial Expression is one that contains two terms.

Trinomial: Trinomials are expressions with three terms.

Polynomial: An Algebraic Expression with two or more terms is referred to as a Polynomial. Students can learn more about these concepts from the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 can be downloaded from the mobile application and website of Extramarks. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 must be downloaded in PDF format for easy access.

After solving Maths Class 7 Chapter 12 Exercise 12.2, students can verify their answers from the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 are a great resource for students to self study. To download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 students can visit the website and mobile application of Extramarks. Moreover, they can download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 in PDF format.

Some other important facts regarding Algebraic expressions are:

  • A Constant has a fixed value while a variable has a wide range of possible values.
  • Basic operations on Variables and Constants can be used to create Algebraic Expressions.
  • Any term in an Algebraic Expression typically has factors. Variables and Numbers are constant factors in Algebra.
  • Any Component of a word is referred to as the coefficient of the term’s remaining portion.
  • A Monomial is the portion of an Algebraic statement that has just one term.
  • A Trinomial contains three terms, whereas a Binomial is a Polynomial with two terms.
  • Similar concepts are referred to as like terms, whereas terms with distinct Algebraic factors are referred to as unlike terms.
  • A term with a Coefficient equal to the total (or difference) of the Coefficients of the two similar terms is equivalent to the sum (or difference) of two like terms.
  • Algebraic Expressions can be used to write rules and formulas in Mathematics in a clear and comprehensive manner.

Exercise 12.2:  The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 will benefit students a lot while solving NCERT Class 7 Maths Chapter 12 Exercise 12.2. Students will find the most accurate answers of all the questions in the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. They can trust the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 as all the resources offered by Extramarks are regularly proofread and updated. The NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 will make studying easier for students. They will find the most simple yet accurate solutions for the exercise in the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2.

Therefore, students must download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2. They are advised to download the NCERT Solutions Class 7 Maths Chapter 12 Exercise 12.2 in PDF format for easy access.


Simplify combining like terms:(i) 21b32+7b20b(ii) z2+13z25z+7z315z(iii) p(pq)q(qp)(iv) 3a2bab(ab+ab)+3ab+ba(v) 5x2y5x2+3yx23y2+x2y2+8xy23y2(vi)(3y2+5y4)(8yy24)


(i) 21b32+7b20b=21b+7b20b32=8b32(ii) z2+13z25z+7z315z=7z3+12z220z(iii) p(pq)q(qp)=pp+qqq+p=pq

(iv) 3a2bab(ab+ab)+3ab+ba=3a2baba+bab+3ab+ba=3aaa2b+b+b+3ababab=a+ab(v) 5x2y5x2+3yx23y2+x2y2+8xy23y2=5x2y+3x2y5x2+x23y2y23y2+8xy2=8x2y4x27y2+8xy2(vi) (3y2+5y4)(8yy24)=3y2+y2+5y8y4+4=4y23y


Add:(i) 3mn,5mn,8mn,4mn(ii) t8tz,3tzz,zt(iii) 7mn+5,12mn+2,9mn8,2mn3(iv) a+b3,ba+3,ab+3(v) 14x+10y12xy13,187x10y+8xy,4xy(vi) 5m7n,3n4m+2,2m3mn5(vii) 4x2y,3xy2,5xy2,5x2y

(viii) 3p2q24pq+5,10p2q2,15+9pq+7p2q2(ix) ab4a,4bab,4a4b(x) x2y21,y21x2,1x2y2


(i) 3mn+( 5mn )+8mn+( 4mn ) =11mn9mn =2mn (ii) ( t8tz )+( 3tzz )+( zt ) = t 8tz+3tz z + z t =5tz (iii) ( 7mn+5 )+( 12mn+2 )+( 9mn8 )+( 2mn3 ) +7mn+5+12mn+2+9mn82mn3 =12mn4 ( iv ) ( a+b3 )+( ba+3 )+( ab+3 ) = a + b 3 +b a + 3 +a b +3 =a+b+3 ( vi ) ( 5m7n )+( 3n4m+2 )+( 2m3mn5 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaeHbwvMCKfMBHbacgaGaa8hkaiaa=LgacaWFPaaabaGaa83maiaa=1gacaWFUbGaey4kaSYaaeWaaeaacqGHsislcaWF1aGaa8xBaiaa=5gaaiaawIcacaGLPaaacqGHRaWkcaWF4aGaa8xBaiaa=5gacqGHRaWkdaqadaqaaiabgkHiTiaa=rdacaWFTbGaa8NBaaGaayjkaiaawMcaaaqaaiabg2da9iaa=fdacaWFXaGaa8xBaiaa=5gacqGHsislcaWF5aGaa8xBaiaa=5gaaeaacqGH9aqpcaWFYaGaa8xBaiaa=5gaaeaacaWFOaGaa8xAaiaa=LgacaWFPaaabaWaaeWaaeaacaWF0bGaeyOeI0Iaa8hoaiaa=rhacaWF6baacaGLOaGaayzkaaGaey4kaSYaaeWaaeaacaWFZaGaa8hDaiaa=PhacqGHsislcaWF6baacaGLOaGaayzkaaGaey4kaSYaaeWaaeaacaWF6bGaeyOeI0Iaa8hDaaGaayjkaiaawMcaaaqaaiabg2da9maaKiaabaGaa8hDaaaacqGHsislcaWF4aGaa8hDaiaa=PhacqGHRaWkcaWFZaGaa8hDaiaa=PhacqGHsisldaajcaqaaiaa=PhaaaGaey4kaSYaaqIaaeaacaWF6baaaiabgkHiTmaaKiaabaGaa8hDaaaaaeaacqGH9aqpcqGHsislcaWF1aGaa8hDaiaa=PhaaeaacaWFOaGaa8xAaiaa=LgacaWFPbGaa8xkaaqaamaabmaabaGaeyOeI0Iaa83naiaa=1gacaWFUbGaey4kaSIaa8xnaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaa8xmaiaa=jdacaWFTbGaa8NBaiabgUcaRiaa=jdaaiaawIcacaGLPaaacqGHRaWkdaqadaqaaiaa=LdacaWFTbGaa8NBaiabgkHiTiaa=HdaaiaawIcacaGLPaaacqGHRaWkdaqadaqaaiabgkHiTiaa=jdacaWFTbGaa8NBaiabgkHiTiaa=ndaaiaawIcacaGLPaaaaeaacaWFRaGaeyOeI0Iaa83naiaa=1gacaWFUbGaey4kaSIaa8xnaiabgUcaRiaa=fdacaWFYaGaa8xBaiaa=5gacqGHRaWkcaWFYaGaey4kaSIaa8xoaiaa=1gacaWFUbGaeyOeI0Iaa8hoaiabgkHiTiaa=jdacaWFTbGaa8NBaiabgkHiTiaa=ndaaeaacqGH9aqpcaWFXaGaa8Nmaiaa=1gacaWFUbGaeyOeI0Iaa8hnaaqaamaabmaabaGaa8xAaiaa=zhaaiaawIcacaGLPaaaaeaadaqadaqaaiaa=fgacqGHRaWkcaWFIbGaeyOeI0Iaa83maaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaa8NyaiabgkHiTiaa=fgacqGHRaWkcaWFZaaacaGLOaGaayzkaaGaa83kamaabmaabaGaa8xyaiabgkHiTiaa=jgacqGHRaWkcaWFZaaacaGLOaGaayzkaaaabaGaeyypa0ZaaqIaaeaacaWFHbaaaiabgUcaRmaaKiaabaGaa8NyaaaacqGHsisldaajcaqaaiaa=ndaaaGaey4kaSIaa8NyaiabgkHiTmaaKiaabaGaa8xyaaaacqGHRaWkdaajcaqaaiaa=ndaaaGaey4kaSIaa8xyaiabgkHiTmaaKiaabaGaa8NyaaaacqGHRaWkcaWFZaaabaGaeyypa0Jaa8xyaiabgUcaRiaa=jgacqGHRaWkcaWFZaaabaWaaeWaaeaacaWF2bGaa8xAaaGaayjkaiaawMcaaaqaamaabmaabaGaa8xnaiaa=1gacqGHsislcaWF3aGaa8NBaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaa83maiaa=5gacqGHsislcaWF0aGaa8xBaiabgUcaRiaa=jdaaiaawIcacaGLPaaacqGHRaWkdaqadaqaaiaa=jdacaWFTbGaeyOeI0Iaa83maiaa=1gacaWFUbGaeyOeI0Iaa8xnaaGaayjkaiaawMcaaaaaaa@10C3@

=5m7n+3n4m+2+2m3mn5 =3m4n3mn3 ( vii ) 4 x 2 y3x y 2 5x y 2 +5 x 2 y =4 x 2 y+5 x 2 y3x y 2 5x y 2 =9 x 2 y8x y 2 ( viii ) ( 3 p 2 q 2 4pq+5 )+( 10 p 2 q 2 )+( 15+pq+7 p 2 q 2 ) =3 p 2 q 2 10 p 2 q 2 +7 p 2 q 2 4pq+9pq+5+15 =5pq+20 ( ix ) ( ab4a )+( 4bab )+( 4a4b ) =abab4a+4a+4b4b =0 ( x ) ( x 2 y 2 1 )+( y 2 1 x 2 )+( 1 x 2 y 2 ) = x 2 y 2 1 + y 2 1 x 2 +1 x 2 y 2 = x 2 y 2 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiabg2da9eHbwvMCKfMBHbacgaGaa8xnaiaa=1gacqGHsislcaWF3aGaa8NBaiabgUcaRiaa=ndacaWFUbGaeyOeI0Iaa8hnaiaa=1gacqGHRaWkcaWFYaGaey4kaSIaa8Nmaiaa=1gacqGHsislcaWFZaGaa8xBaiaa=5gacqGHsislcaWF1aaabaGaeyypa0Jaa83maiaa=1gacqGHsislcaWF0aGaa8NBaiabgkHiTiaa=ndacaWFTbGaa8NBaiabgkHiTiaa=ndaaeaadaqadaqaaiaa=zhacaWFPbGaa8xAaaGaayjkaiaawMcaaaqaaiaa=rdacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8xEaiabgkHiTiaa=ndacaWF4bGaa8xEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=vdacaWF4bGaa8xEamaaCaaaleqabaGaa8NmaaaakiabgUcaRiaa=vdacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8xEaaqaaiaa=1dacaWF0aGaa8hEamaaCaaaleqabaGaa8Nmaaaakiaa=LhacqGHRaWkcaWF1aGaa8hEamaaCaaaleqabaGaa8Nmaaaakiaa=LhacqGHsislcaWFZaGaa8hEaiaa=LhadaahaaWcbeqaaiaa=jdaaaGccqGHsislcaWF1aGaa8hEaiaa=LhadaahaaWcbeqaaiaa=jdaaaaakeaacaWF9aGaa8xoaiaa=HhadaahaaWcbeqaaiaa=jdaaaGccaWF5bGaeyOeI0Iaa8hoaiaa=HhacaWF5bWaaWbaaSqabeaacaWFYaaaaaGcbaWaaeWaaeaacaWF2bGaa8xAaiaa=LgacaWFPbaacaGLOaGaayzkaaaabaWaaeWaaeaacaWFZaGaa8hCamaaCaaaleqabaGaa8Nmaaaakiaa=fhadaahaaWcbeqaaiaa=jdaaaGccqGHsislcaWF0aGaa8hCaiaa=fhacqGHRaWkcaWF1aaacaGLOaGaayzkaaGaey4kaSYaaeWaaeaacqGHsislcaWFXaGaa8hmaiaa=bhadaahaaWcbeqaaiaa=jdaaaGccaWFXbWaaWbaaSqabeaacaWFYaaaaaGccaGLOaGaayzkaaGaey4kaSYaaeWaaeaacaWFXaGaa8xnaiabgUcaRiaa=bhacaWFXbGaey4kaSIaa83naiaa=bhadaahaaWcbeqaaiaa=jdaaaGccaWFXbWaaWbaaSqabeaacaWFYaaaaaGccaGLOaGaayzkaaaabaGaeyypa0Jaa83maiaa=bhadaahaaWcbeqaaiaa=jdaaaGccaWFXbWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8xmaiaa=bdacaWFWbWaaWbaaSqabeaacaWFYaaaaOGaa8xCamaaCaaaleqabaGaa8NmaaaakiabgUcaRiaa=DdacaWFWbWaaWbaaSqabeaacaWFYaaaaOGaa8xCamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=rdacaWFWbGaa8xCaiabgUcaRiaa=LdacaWFWbGaa8xCaiabgUcaRiaa=vdacqGHRaWkcaWFXaGaa8xnaaqaaiabg2da9iaa=vdacaWFWbGaa8xCaiaa=TcacaWFYaGaa8hmaaqaamaabmaabaGaa8xAaiaa=HhaaiaawIcacaGLPaaaaeaadaqadaqaaiaa=fgacaWFIbGaa8xlaiaa=rdacaWFHbaacaGLOaGaayzkaaGaa83kamaabmaabaGaa8hnaiaa=jgacaWFTaGaa8xyaiaa=jgaaiaawIcacaGLPaaacaWFRaWaaeWaaeaacaWF0aGaa8xyaiaa=1cacaWF0aGaa8NyaaGaayjkaiaawMcaaaqaaiabg2da9iaa=fgacaWFIbGaeyOeI0Iaa8xyaiaa=jgacqGHsislcaWF0aGaa8xyaiabgUcaRiaa=rdacaWFHbGaey4kaSIaa8hnaiaa=jgacqGHsislcaWF0aGaa8Nyaaqaaiabg2da9iaa=bdaaeaadaqadaqaaiaa=HhaaiaawIcacaGLPaaaaeaadaqadaqaaiaa=HhadaahaaWcbeqaaiaa=jdaaaGccqGHsislcaWF5bWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8xmaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaa8xEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=fdacqGHsislcaWF4bWaaWbaaSqabeaacaWFYaaaaaGccaGLOaGaayzkaaGaey4kaSYaaeWaaeaacaWFXaGaeyOeI0Iaa8hEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=LhadaahaaWcbeqaaiaa=jdaaaaakiaawIcacaGLPaaaaeaacqGH9aqpdaajcaqaaiaa=HhadaahaaWcbeqaaiaa=jdaaaaaaOGaeyOeI0YaaqIaaeaacaWF5bWaaWbaaSqabeaacaWFYaaaaaaakiabgkHiTmaaKiaabaGaa8xmaaaacqGHRaWkdaajcaqaaiaa=LhadaahaaWcbeqaaiaa=jdaaaaaaOGaeyOeI0YaaqIaaeaacaWFXaaaaiabgkHiTmaaKiaabaGaa8hEamaaCaaaleqabaGaa8NmaaaaaaGccqGHRaWkcaWFXaGaeyOeI0Iaa8hEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=LhadaahaaWcbeqaaiaa=jdaaaaakeaacqGH9aqpcqGHsislcaWF4bWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8xEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=fdaaaaa@3A6A@


Subtract:(i) 5y2fromy2(ii) 6xyfrom12xy(iii) (ab)from(a+b)(iv) a(b5)fromb(5a)(v) m2+5mnfrom4m23mn+8(vi) x2+10x5from5x10(vii) 5a27ab+5b2from3ab2a22b2(viii) 4pq5q23p2from5p2+3q2pq


(i) y2(5y2)=6y2(ii) 12xy(6xy)=18xy(iii) (a+b)(ab)=a+ba+b=2b(iv) b(5a)a(b5)=5babab+5a=5a+5b2ab

(v) (4m23mn+8)(m2+5mn)=4m23mn+8+m25mn=5m28mn+8(vi) (5x10)(x2+10x5)=5x10+x210x+5=x25x5(vii) (3ab2a22b2)(5a27ab+5b2)=3ab2a22b25a2+7ab5b2=10ab7a27b2(viii) (5p2+3q2pq)(4pq5q23p2)=5p2+3q2pq4pq5q23p2=8p2+8q25pq


(a) What should be added to x2+xy+y2 to obtain 2x2+3xy?(b) What should be subtracted from 2a+8b+10 to get 3a+7b+16?


(a) Let a be the required sum. Then, we have a+( x 2 +xy+y 2 )=( 2x 2 +3xy ) a= 2x 2 +3xy( x 2 +xy+ y 2 ) = 2x 2 +3xy x 2 xy y 2 = x 2 y 2 +2xy (b) Let x be the required sum. Then, we have 2a+8b+10p=3a+7b+16 p=2a+8b+10(3a+7b+16) =2a+8b+10+3a7b16 =5a+b6 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5137@


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


Let p be the required term. (3 x 2 4 y 2 + 5xy + 20) p = x 2 y 2 + 6xy + 20 p = (3 x 2 4 y 2 + 5xy + 20) ( x 2 y 2 + 6xy + 20) = 3 x 2 4 y 2 + 5xy + 20 + x 2 + y 2 6xy 20 = 3 x 2 + x 2 4 y 2 + y 2 + 5xy 6xy + 20 20 = 4 x 2 3 y 2 xy MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfKttLearuGu1bxzLbIrVjxyKLwyUbqeduuDJXwAKbYu51MyVXgatCvAUfeBSjuyZL2yd9gzLbvyNv2CaeHbd9wDYLwzYbItLDharyavP1wzZbItLDhis9wBH5garqqr1ngBPrgifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaeHbwvMCKfMBHbacgaGaa8htaiaa=vgacaWF0bGaa8hOaiaa=bhacaWFGcGaa8Nyaiaa=vgacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=jhacaWFLbGaa8xCaiaa=vhacaWFPbGaa8NCaiaa=vgacaWFKbGaa8hiaiaa=rhacaWFLbGaa8NCaiaa=1gacaWFUaaabaGaa8hkaiaa=ndacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8hiaiaa=rdacaWF5bWaaWbaaSqabeaacaWFYaaaaOGaa8hOaiaa=TcacaWFGaGaa8xnaiaa=HhacaWF5bGaa8hOaiaa=TcacaWFGaGaa8Nmaiaa=bdacaWFPaGaa8hiaiabgkHiTiaa=bkacaWFWbGaa8hOaiabg2da9iabgkHiTiaa=bkacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8hOaiaa=LhadaahaaWcbeqaaiaa=jdaaaGccaWFGcGaa83kaiaa=bcacaWF2aGaa8hEaiaa=LhacaWFGcGaa83kaiaa=bcacaWFYaGaa8hmaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caWFWbGaaGPaVlaa=bkacqGH9aqpcaWFGaGaa8hkaiaa=ndacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8hOaiabgkHiTiaa=bcacaWF0aGaa8xEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacaWFRaGaa8hiaiaa=vdacaWF4bGaa8xEaiaa=bkacaWFRaGaa8hiaiaa=jdacaWFWaGaa8xkaiaa=bcaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaeyOeI0Iaa8hiaiaa=HcacqGHsislcaWFGcGaa8hEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacqGHsislcaWFGcGaa8xEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacaWFRaGaa8hiaiaa=zdacaWF4bGaa8xEaiaa=bkacaWFRaGaa8hiaiaa=jdacaWFWaGaa8xkaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iaa=bcacaWFZaGaa8hEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacqGHsislcaWFGaGaa8hnaiaa=LhadaahaaWcbeqaaiaa=jdaaaGccaWFGcGaa83kaiaa=bcacaWF1aGaa8hEaiaa=LhacaWFGcGaa83kaiaa=bcacaWFYaGaa8hmaiaa=bcacaWFRaGaa8hOaiaa=HhadaahaaWcbeqaaiaa=jdaaaaakeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaa8hOaiaa=TcacaWFGcGaa8xEamaaCaaaleqabaGaa8NmaaaakiabgkHiTiaa=zdacaWF4bGaa8xEaiaa=bkacqGHsislcaWFYaGaa8hmaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iaa=bcacaWFZaGaa8hEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacaWFRaGaa8hOaiaa=HhadaahaaWcbeqaaiaa=jdacaWFGcaaaOGaeyOeI0Iaa8hiaiaa=rdacaWF5bWaaWbaaSqabeaacaWFYaaaaOGaa8hOaiaa=TcacaWFGcGaa8xEamaaCaaaleqabaGaa8Nmaaaakiaa=bkacaWFRaGaa8hiaiaa=vdacaWF4bGaa8xEaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaacqGHsislcaWFGaGaa8Nnaiaa=HhacaWF5bGaa8hOaiaa=TcacaWFGaGaa8Nmaiaa=bdacqGHsislcaWFGaGaa8Nmaiaa=bdaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpcaWFGaGaa8hnaiaa=HhadaahaaWcbeqaaiaa=jdaaaGccqGHsislcaWFGaGaa83maiaa=LhadaahaaWcbeqaaiaa=jdaaaGccqGHsislcaWFGcGaa8hEaiaa=Lhaaaaa@3382@


(a) From the sum of 3xy+11 and y11, subtract 3xy11.(b) From the sum of 4+3 x and 54x+2x2, subtract the sum of 3x25x and x2+2x+5.


(a) (3x y +11)+( y 11)=3xy +11y 11=3x y y +1111=3x2y(3x 2y)(3x y 11)=3x 2y3x + y +11=3x 3x 2y + y +11= y +11(b) (4+3x)+(54x +2x2)=4+3x +54x +2x2=3x 4x +2x2 +4+5=x +2x2 +9(3x25x)+(x2 +2x +5)=3x25x x2 +2x +5=3x2 x2 5x +2x +5=2x23x +5(x +2x2 +9)(2x23x +5)=x +2x2 +92x2 +3x5= x +3x +2x22x2 +95=2x +4

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