Ex 12.2 Q3 Algebraic-Expressions Solutions NCERT Maths Class 7

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Question

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)

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

 Video Solution
Algebraic Expressions
Ex 12.2 | Question 3

Text 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)

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

  
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