Exercise 12.2 Algebraic-Expressions -NCERT Solutions Class 7

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Chapter 12 Ex.12.2 Question 1

Simplify combining like terms:

(i)

\(21b \,– 32 + 7b – 20b \)

(ii)

\(– z2 + 13z2 – 5z + 7z3 – 15z\)

(iii)

\(p – (p – q) – q – (q – p)\)

(iv)

\(3a – 2b – ab – (a – b + ab) + 3ab + b – a \)

(v)

\(\begin{align}&\!5x^2y\!-\!5x^2 \!\!+\! \! 3yx^2 \!–\! 3y^2 \! \!+\! \!\\ &x^2 \!–\! y^2 \!\!+\! \!8xy^2 \!–\! 3y^2\end{align}\)

(vi)

\((3y2 + 5y – 4) – (8y – y2 – 4)\)

Solution

Video Solution

What is Known?

Like Terms

What is unknown?

How to simplify Like Terms

Reasoning:

This is based on concept identifying like terms and then performing the arithmetic operation of like terms to simplify them.

Steps:

(i)

\[\begin{align}&{21b-32 + 7b-20b}\\&= 21b + 7b - 20b - 32\\&= 8b - 32\end{align}\]

(ii)

\[\begin{align}&{-{z^2} + 13{z^2}-5z + 7{z^3}-15z}\\&= {7{z^3} + 12{z^2} - 20z}\end{align}\]

(iii)

\[\begin{align}&{p-\left( {p-q} \right)-q-\left( {q-p} \right)}\\&= {\rm{{ }}p-p + q-q-q + p}\\&= {\rm{{ }}p - q}\end{align}\]

(iv)

\[\begin{align}&{3a\!-\!2b\!-\!ab\!-\!\left( {a\!-\!b \!+\! ab} \right) \!+\! 3ab \!+\! b\!-\!a}\\&\!=\! {3a\!-\!2b\!-\!ab\!-\!a \!+\! b ab \!+\! 3ab \!+\! b\!-\!a}\\&\!=\! {3a\!-\!a\!-\!a\!-\!2b \!+\! b \!+\! b \!-\! ab\!-\!ab \!+\! 3ab}\\&\!=\! {a \!+\! ab}\end{align}\]

(v)

\[\begin{align} & \left[\begin{array} & 5{{x}^{2}}y-5{{x}^{2}}+3y{{x}^{2}}-3{{y}^{2}}+ \\ {{x}^{2}}-{{y}^{2}}+8x{{y}^{2}}-3{{y}^{2}} \\ \end{array} \right] \\ & =\left[\begin{array} & 5{{x}^{2}}y+3y{{x}^{2}}-3{{y}^{2}}-{{y}^{2}}+ \\ {{x}^{2}}-5{{x}^{2}}+8x{{y}^{2}}-3{{y}^{2}} \\ \end{array} \right] \\ & =8{{x}^{2}}y-7{{y}^{2}}-4{{x}^{2}}+8x{{y}^{2}} \\ \end{align}\]

(vi)

\[\begin{align}&{\left( {3{y^2} + 5y-4} \right)-\left( {8y-{y^2}-4} \right)}\\& = {\rm{{ }}3{y^2} + 5y--4-8y + {y^2} + 4}\\& = {\rm{{ }}3{y^2} + {y^2} + 5y-8y + 4-4}\\&= {\rm{{ }}4{y^2}-3y}\end{align}\]

Chapter 12 Ex.12.2 Question 2

Add:

(i) \(3\,mn, – 5\,mn, 8\,mn, – 4\,mn \)

(ii) \(t – 8tz, 3tz – z, z – t\)

(iii) \(– 7\,mn + 5, 12\,mn + 2, \\9\,mn – 8, – 2\,mn – 3 \)

(iv) \(a + b – 3, b – a + 3, a – b + 3\)

(v) \(14x + 10y – 12xy – 13,\\18 – 7x – 10y + 8xy, 4xy \)

(vi) \(5\,m – 7\,n, 3\,n – 4\,m + 2, 2\,m – 3\,mn – 5\)

(vii) \(4x^2y, – 3xy^2, –5xy^2, 5x^2y \)

(viii) \(3p^2q^2 – 4pq + 5, – 10p^2q^2,\\ 15 + 9pq + 7p^2q^2\)

(ix) \(ab – 4a, 4b – ab, 4a – 4b\)

(x) \(x^2 – y^2 – 1, y^2 – 1 – x^2, 1 – x^2 – y^2\)

Solution

Video Solution

What is Known?

Like Terms

What is unknown?

How to add or subtract Like Terms

Reasoning:

This is based on concept identifying like terms and then performing the addition operation of like terms.

Steps:

(i)

\[\begin{align}&{3mn,-5mn,8mn,-4mn}\\&\!=\! {3mn \!+\! \left( {-5mn} \right) \!+\! 8mn \!+\! \left( {-4mn} \right)}\\&\!=\! {11mn\!-\!9mn}\\&\!=\! 2mn\end{align}\]

(ii)

\[\begin{align}&{t-8tz,3tz-z,z-t}\\&= {t-8tz + 3tz-z + z-t}\\&= { - 5tz}\end{align}\]

(iii)

\[\begin{align}&{ \!- \! 7mn \!+ \! 5, \!12mn \!+ \!2,9mn \!- \!8,-\!2mn\!-\!3}\\&\!\!= { \!\!-\!7mn \!+ \! 5 \! + \! \!12mn \!+ \! 2 \! + \! 9mn \!\!- \!8\! \!+ \! \! \left( { \!\!- \!2mn} \right) \!\!- \!\!3}\\&\!\!= \!{ \!-\!7mn \!+ \!5 \! + \! 12mn \!+ \! 2 \! + \! 9mn \!- \!8 \!- \!2mn \!- \!\!3}\\&\!\! =\! {12mn \!- \!4}\end{align}\]

(iv)

\[\begin{align}&{a \!+\! b\!-\!3,b\!-\!a \!+\! 3,a\!-\!b \!+\! 3}\\&=\! a \!+\! b\!-\!3 \!+\! b\!-\!a \!+\! 3 \!+\! a\!-\!b \!+\! 3\\&= \!a \!+\! b \!+\! 3\end{align}\]

(v)

\[\begin{align}&{14x \!+\! 10y\!-\!12xy\!-\!13,18\!-\!7x\!-\!10y \!+\! 8xy,4xy}\\&= \!14x \!+ \!10y \!- \!12xy \!- \!13 \!+ \! 18 \!- \!7x \!- \!10y \! + \!8xy \!+ \!4xy\\& = \!7x \!+\! 5\end{align}\]

(vi)

\[\begin{align}&{5m\!-\!7n,3n\!-\!4m \!+\! 2,2m\!-\!3mn\!-\!5}\\&\!= \!5m\!-\!7n \!+\! 3n\!-\!4m \!+\! 2 \!+\! 2m\!-\!3mn\!-\!5\\&\!=\! 5m \!-\! 4m \!+\! 2m\!-\!7n \!+\! 3n \!+\! 2\!-\!5 \!-\! 3mn\\&\!=\! 3m\!-\!4n\!-\!3mn\! -\! 3\end{align}\]

(vii)

\[\begin{align}&{4{x^2}y,-3x{y^2},-5x{y^2},5{x^2}y}\\&= \!4{x^2}y \!+\! \left( {-3x{y^2}} \right) \!+\! \left( {-5x{y^2}} \right) \!+\! 5{x^2}y\\&= \!4{x^2}y\!-\!3x{y^2}\!-\!5x{y^2} \!+\! 5{x^2}y\\&= \!9{x^2}y\!-\!8x{y^2}\end{align}\]

(viii) 

\[\begin{align}&{3{p^2}{q^2}\!\!-4pq\! + \!5,\!\!\!-\!\!10{p^2}{q^2}\!,\!15 \!\!+ \!9pq + 7{p^2}{q^2}}\\&\!\!=\! 3{p^2}{q^2}\!\!-\!4pq \!+\! 5\!-\!\!10{p^2}{q^2} \!\!+ \!\!15 \!+ \!9pq\! +\! 7{p^2}{q^2}\\&\!\!=\! 3{p^2}{q^2}\!\! +\!\! 7{p^2}{q^2}\!\!-\!\!10{p^2}{q^2}\!\!-\!4pq\! +\! 9pq \!+\! 5\! +\! 15\\& \!\!=\! 5pq + 20\end{align}\]

(ix)

\[\begin{align}&{ab-4a,4b-ab,4a-4b}\\& = ab-4a + 4b-ab + 4a-4b\\&= 0\end{align}\]

(x)

\[\begin{align}&{{x^2}\!-\!{y^2}\!-\!1,{y^2}\!-\!1\!-\!{x^2},1\!-\!{x^2}\!-\!{y^2}}\\& =\! {x^2}\!-\!{y^2}\!-\!1\!\! + \!{y^2}\!\!-\!1\!-\!{x^2} \!+ \!1\!-\!{x^2}\!-\!{y^2}\\&=\!\! - \!{x^2} \!-\! {y^2} \!-\! 1\end{align}\]

Chapter 12 Ex.12.2 Question 3

Subtract:

(i)

\(-5y^2 \text{ from } y^2 \)

(ii)

\(6xy \text{ from } –12xy\)

(iii)

\((a – b) \text{ from } (a + b)\)

(iv)

\(a(b – 5) \text{ from } b (5 – a)\)

(v)

\(-m^2 + 5mn \text{ from } 4m^2 – 3mn + 8 \)

(vi)

\(-x^2 + 10x – 5 \text{ from } 5x – 10\)

(vii)

\(5a^2 – 7ab + 5b^2 \text{ from } 3ab – 2a^2 – 2b^2\)

(viii)

\(4pq – 5q^2 – 3p^2 \text{ from } 5p^2 + 3q^2 – pq\)

Solution

Video Solution

What is known?

Like Terms

Reasoning:

This is based on concept identifying like terms and then performing the subtraction operation of like terms.

Steps:

(i) 

\[\begin{align}&-5{y^2}\,\text{from}\,{y^2}\\&= {y^2}-\left( { - 5{y^2}} \right)\\&= {y^2} + 5{y^2}\\&= 6{y^2}\end{align}\]

(ii)

\[\begin{align}&{6xy\,\text{from}\,-12xy}\\&=  - 12xy-6xy\\& =  - 18x\end{align}\]

(iii)

\[\begin{align}&{\left( {a-b} \right)\,\text{from}\,\left( {a + b} \right)}\\&= \left( {a + b} \right) - (a-b)\\& = a + b - a + b\\&= 2b\end{align}\]

(iv)

\[\begin{align}&{a\left( {b-5} \right)\,\text{from}\,b\left( {5-a} \right)}\\&= \rm{{ }}b\left( {5-a} \right)-a\left( {b-5} \right)\\& = 5b - {\rm{ab }}-{\rm{ab }}+{\rm{5a }}  \\&= {\rm{ 5a}} + {\rm{5b }} - {\rm{2ab }}
\end{align}\]

(v)

\[\begin{align}&{-{m^2} + 5mn\,\text{from}\,4{m^2}-3mn + 8}\\&= 4{m^2}-3mn \!+\! 8\!-\!(-{m^2} + 5mn)\\&= 4{m^2}-3mn + 8 + {m^2} \!-\! 5mn\\&= 5{m^{2-}}8mn + 8 \end{align}\]

(vi)

\[\begin{align}&{ - {x^2} + 10x-5\,\text{from}\,5x-10}\\&= 5x-10-\left( {-{x^2} + 10x-5} \right)\\& = 5x-10 + {x^2} - 10x + 5\\&= {x^2}-5x - 5\end{align}\]

(vii)

\[\begin{align}&5{a^2}\!-\!7ab \!+\! 5{b^2}\,\text{ from }\,3ab\!-\!2{a^2}\!-\!2{b^2}\\&= \!3ab\!-\!2{a^2}\!-\!2{b^2}\!-\!\left( {5{a^2}\!-\!7ab \!+\! 5{b^2}} \right)\\& =\! 3ab\!-\!2{a^2}\!-\!2{b^2}\!-\!5{a^2} \!+\! 7ab \!-\! 5{b^2}\\&=\! 10ab\!-\!{a^2}\!-\!7{b^2}\end{align}\]

(viii)

\[\begin{align}&{4pq\!-\!5{q^2}\!-\!3{p^2}\,\text{ from }\,5{p^2} \!+\! 3{q^2}\!-\!pq}\\&= \!5{p^2} \!+\! 3{q^2}\!-\!pq\!-\!(4pq\!-\!5{q^2}\!-\!3{p^2})\\&= \!5{p^2} \!+\! 3{q^2}\!-\!pq\!-\!4pq \!+\! 5{q^2} \!+\! 3{p^2}\\&= \!8{p^2} \!+\! 8{q^2}\!-\!5pq\end{align}\]

Chapter 12 Ex.12.2 Question 4

(a) What should be added to \(x^2 + xy + y^2\) to obtain \(2x^2 + 3xy?\)

(b) What should be subtracted from \(2a + 8b + 10\) to get \(– 3a + 7b + 16? \)

Solution

Video Solution

What is known?

We know that arithmetic operation will be applied.

Reasoning:

In this question basic concept of arithmetic operations is applied. We are given two terms and asked what should be added to one term to get the required answer. For this we will subtract the first term from the answer to get what should be added. E.g. what should be added to \(3\) to get \(5.\) We will subtract \(3\) from \(5.\)

Steps:

1st term \(=x^2 + xy + y^2\)

Answer term \(= 2x^2 + 3xy\)

2nd term \(=\) Answer term \(-\) 1st term

2nd term

\[\begin{align}&= 2x^2 + 3xy – (x^2 + xy + y^2)\\&= 2{x^2} + {\rm{ }}3xy{\rm{ }}-{\rm{ }}{x^2} - xy{\rm{ }} - {\rm{ }}{y^2}\\&= {x^{2{\rm{ }} + }}2xy{\rm{ }} - {y^2}\end{align}\]

So, \(x^2 + 2xy - y^2\) should be added to \(x^2 + xy + y^2\) to obtain \(2x^2 + 3xy\)

(b) What should be subtracted from \(2a + 8b + 10\) to get \(–3a + 7b + 16?\)

What is known?

We know that arithmetic operation will be applied.

Reasoning:

In this question basic concept of arithmetic operations is applied. We are given two terms and asked what should be subtracted from one term to get the required answer. For this we will subtract the given answer from the 1st term to get what should be subtracted. E.g: what should be subtracted from \(5\) to get \(3.\) We will subtract \(3\) from \(5. \)

Steps:

1st term \(= 2a + 8b + 10 \)

Answer term \(= \,– 3a + 7b + 16\)

2nd term = 1st term - Answer

2nd term

\[\begin{align}&= \rm{}2a + 8b + 10 – (– 3a + 7b + 16)\\&= {\rm{ }}2a + {\rm{ }}8b + {\rm{ }}10{\rm{ }} + {\rm{ }}3a - {\rm{ }}7b - {\rm{ }}16\\&= 5a{\rm{ }} + {\rm{ }}b{\rm{ }} - {\rm{ }}6\end{align}\]

So, \(5a + b – 6\) should be subtracted from \(2a + 8b + 10\) to obtain \(– 3a + 7b + 16\)

Chapter 12 Ex.12.2 Question 5

What should be taken away from \(3x^2 – 4y^2 + 5xy + 20\)  to obtain  \(– x^2 – y^2 + 6xy + 20?\)

Solution

Video Solution

What is known?

we know that arithmetic operation will be applied.

Reasoning:

In this question basic concept of arithmetic operations is applied. We are given two terms and asked what should be subtracted from one term to get the required answer. For this we will subtract the given answer from the 1st term to get what should be subtracted. E.g: what should be subtracted from \(5\) to get \(3.\) We will subtract \(3\) from \(5.\)

Steps:

1st term \(= 3x^2 – 4y^2 + 5xy + 20\)

Answer term \(= \, – x^2 – y^2 + 6xy + 20\)

2nd term = 1st term – Answer

2nd term

\[\begin{align}&\!\!\!\!\!\!= \!3x^2\! –\! 4y^2 \!+\! 5xy \!+ \!20 \!–\! (\!– \!x^2\! –\! y^2\! + \!6xy \!+ \!20)\\&\!\!\!\!\!\!= \!3{x^2}\!\!-\!4{y^2} \!+ \!5xy \!+\! 20 \!+\! {x^2} \!\!+ \!{y^2} \!-\! 6xy\! -\! 20\\&\!\!\!\!\!\! = \!4{x^2} \!-\! 3{y^2} \!- \!xy\end{align}\]

Chapter 12 Ex.12.2 Question 6

(a) From the sum of \(3x – y + 11 \text{ and } – y – 11,\) subtract \(3x – y – 11.\)

(b) From the sum of \(4 + 3x\) and \(5 – 4x + 2x^2,\) subtract the sum of \(3x^2 – 5x\) and \(-x^2 + 2x + 5\)

Solution

Video Solution

What is known?

Like Terms

Reasoning:

This is based on concept identifying like terms and then performing the arithmetic operation of like terms as given in the question.

Steps:

(a) From the sum of \(3x – y + 11\) and \(– y – 11\), subtract \(3x – y – 11.\)

First, we add \(3x – y + 11\) and \(– y – 11\)

\[\begin{align}&= {\rm{ }}3x-y + {\rm{ }}11{\rm{ }} + {\rm{ }}\left( {-y-{\rm{ }}11} \right)\\& = {\rm{ }}3x-y + {\rm{ }}11{\rm{ }}-y-{\rm{ }}11\\&= {\rm{ }}3x{\rm{ }} - 2y\end{align}\]

Now from \(3x – 2y\) subtract \(3x – y – 11\)

\[\begin{align}&= {\rm{ }}3x{\rm{ }}-{\rm{ }}2y{\rm{ }}-{\rm{ }}\left( {3x-y-{\rm{ }}11} \right)\\&= {\rm{ }}3x{\rm{ }}-{\rm{ }}2y{\rm{ }}-{\rm{ }}3x + y + {\rm{ }}11\\&= {\rm{ }} - {\rm{ }}y{\rm{ }} + {\rm{ }}11
\end{align}\]

(b) From the sum of \(4 + 3x\) and \(5 – 4x + 2x^2\), subtract the sum of \(3x^2 – 5x \) and  \(–x^2 + 2x + 5 \)

Step 1 \(=\) First, add \(4 + 3x\) and \(5 – 4x + 2x^2\)

Step 2 \(=\) Then, add \(3x^2 – 5x\) and \(–x^2 + 2x + 5\)

Step 3 \(=\) Subtract the resultant in step \(2\) from resultant of step \(1\)

Steps:

Add \(4 + 3x\) and \(5 – 4x + 2x^2\)

\[\begin{align}&= {\rm{ }}4{\rm{ }} + {\rm{ }}3x + {\rm{ }}5{\rm{ }}--{\rm{ }}4x + {\rm{ }}2{x^2}\\&= {\rm{ }}2{x^2} - x{\rm{ }} + {\rm{ }}9\end{align}\]

Now add \(3x^2 – 5x\) and \(–x^2 + 2x + 5\)

\[\begin{align}&= {\rm{ }}3{x^2}-{\rm{ }}5x + \left( {{\rm{ }}-{x^2} + {\rm{ }}2x + {\rm{ }}5} \right)\\& = {\rm{ }}3{x^2}-{\rm{ }}5x--{x^2} + {\rm{ }}2x + {\rm{ }}5\\& = {\rm{ }}2{x^2}-{\rm{ }}3x{\rm{ }} + {\rm{ }}5\end{align}\]

Now subtract \(2x^2 – 3x + 5\) from \(2x^2 -x + 9 \)

i.e.

\[\begin{align}&{2{x^2} - x+ 9-\left( {2{x^2}-3x + 5} \right)}\\&=2{x^2} - x +9-2{x^2} + 3x-5\\&= 2x + 4\end{align}\]

  
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