Ex 12.2 Q6 Algebraic-Expressions Solutions NCERT Maths Class 7

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Question

(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 \(-x^2 + 2x + 5\)

Text 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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