Ex 12.2 Q1 Algebraic-Expressions Solutions NCERT Maths Class 7

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Question

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)

$$(3y2 + 5y – 4) – (8y – y2 – 4)$$

Video Solution
Algebraic Expressions
Ex 12.2 | Question 1

Text 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}

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