Ex 12.2 Q2 Algebraic-Expressions Solutions NCERT Maths Class 7

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Question

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

Text 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)

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

  
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