## NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions (EX 12.2) Exercise 12.2

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## Access NCERT Solutions for Maths Class 7 Chapter 12- Algebraic Expressions

Exercise 12.2

Refer to pages 10-13 for exercise 12.2 in the PDF

1. Simplify combining like terms:

(i) \[21b - 32 + 7b - 20b\]

Ans: Combine the like terms (the terms which have the same variables and power) of variable $b$ and simplify.

$21b - 32 + 7b - 20b = 21b + 7b - 20b - 32$

$21b - 32 + 7b - 20b = 28b - 20b - 32$

$21b - 32 + 7b - 20b = 8b - 32$

The obtained expression is $8b - 32$.

(ii) $ - {z^2} + 13{z^2} - 5z + 7{z^3} - 15z$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${z^2}$ and the variable $z$ and simplify.

$ - {z^2} + 13{z^2} - 5z + 7{z^3} - 15z = 7{z^3} + \left( {13{z^2} - {z^2}} \right) - \left( {5z + 15z} \right)$$ - {z^2} + 13{z^2} - 5z + 7{z^3} - 15z = 7{z^3} + 12{z^2} - 20z$

The obtained expression is $7{z^3} + 12{z^2} - 20z$.

(iii) $p - \left( {p - q} \right) - q - \left( {q - p} \right)$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $p$ and the variable $q$ and simplify.

$p - \left( {p - q} \right) - q - \left( {q - p} \right) = p - p + q - q - q + p$

$p - \left( {p - q} \right) - q - \left( {q - p} \right) = p - p + p + q - q - q$

$p - \left( {p - q} \right) - q - \left( {q - p} \right) = p - q$

The obtained expression is $p - q$.

(iv) $3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $a$, the variable $b$ and the variable $ab$ then simplify.

$3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a = 3a - 2b - ab - a + b - ab + 3ab + b - a$$3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a = 3a - a - a - 2b + b + b - ab - ab + 3ab$$3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a = 2a - a - 2b + 2b - 2ab + 3ab$$3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a = a - 0 + ab$$3a - 2b - ab - \left( {a - b + ab} \right) + 3ab + b - a = a + ab$

The obtained expression is $a + ab$.

(v) $5{x^2}y - 5{x^2} + 3y{x^2} - 3{y^2} + {x^2} - {y^2} + 8x{y^2} - 3{y^2}$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${x^2}$, the variable ${y^2}$ and the variable ${x^2}y$ then simplify.

$5{x^2}y - 5{x^2} + 3y{x^2} - 3{y^2} + {x^2} - {y^2} + 8x{y^2} - 3{y^2} = 5{x^2}y + 3y{x^2} + 8x{y^2} - 5{x^2} + {x^2} - 3{y^2} - {y^2} - 3{y^2}$$5{x^2}y - 5{x^2} + 3y{x^2} - 3{y^2} + {x^2} - {y^2} + 8x{y^2} - 3{y^2} = \left( {5{x^2}y + 3y{x^2}} \right) + 8x{y^2} - \left( {5{x^2} - {x^2}} \right) - \left( {3{y^2} + {y^2} + 3{y^2}} \right)$$5{x^2}y - 5{x^2} + 3y{x^2} - 3{y^2} + {x^2} - {y^2} + 8x{y^2} - 3{y^2} = 8{x^2}y + 8x{y^2} - 4{x^2} - 7{y^2}$

The obtained expression is $8{x^2}y + 8x{y^2} - 4{x^2} - 7{y^2}$.

(vi) $\left( {3{y^2} + 5y - 4} \right) - \left( {8y - {y^2} - 4} \right)$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $y$, the variable ${y^2}$ and the constants then simplify.

$\left( {3{y^2} + 5y - 4} \right) - \left( {8y - {y^2} - 4} \right) = 3{y^2} + 5y - 4 - 8y + {y^2} + 4$$\left( {3{y^2} + 5y - 4} \right) - \left( {8y - {y^2} - 4} \right) = 3{y^2} + {y^2} + 5y - 8y - 4 + 4$$\left( {3{y^2} + 5y - 4} \right) - \left( {8y - {y^2} - 4} \right) = 4{y^2} - 3y - 0$

$\left( {3{y^2} + 5y - 4} \right) - \left( {8y - {y^2} - 4} \right) = 4{y^2} - 3y$

The obtained expression is $4{y^2} - 3y$.

2. Add:

(i) $3mn, - 5mn,8mn, - 4mn$

Ans: $3mn, - 5mn,8mn, - 4mn$ all are like terms (the terms which have the same variables and power) so add the coefficients.

$3mn, - 5mn,8mn, - 4mn = 3mn + \left( { - 5mn} \right) + 8mn + \left( { - 4mn} \right)$$3mn, - 5mn,8mn, - 4mn = \left( {3 - 5 + 8 - 4} \right)mn$

$3mn, - 5mn,8mn, - 4mn = \left( { - 2 + 4} \right)mn$

$3mn, - 5mn,8mn, - 4mn = 2mn$

Therefore, the obtained sum of the given terms are $2mn$.

(ii) $t - 8tz,3tz - z,z - t$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $t$, the variable $z$ and the variable $tz$ then add the coefficients.

$t - 8tz,3tz - z,z - t = t - 8tz + 3tz - z + z - t$

$t - 8tz,3tz - z,z - t = t - t - 8tz + 3tz - z + z$

$t - 8tz,3tz - z,z - t = \left( {1 - 1} \right)t + \left( { - 8 + 3} \right)tz + \left( { - 1 + 1} \right)z$

$t - 8tz,3tz - z,z - t = 0 - 5tz + 0$

$t - 8tz,3tz - z,z - t = - 5tz$

Therefore, the obtained sum of the given terms are $ - 5tz$.

(iii) $ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $mn$ and the constants then add the coefficients.

$ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3 = - 7mn + 5 + 12mn + 2 + 9mn - 8 + \left( { - 2mn} \right) - 3$$ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3 = - 7mn + 12mn + 9mn - 2mn + 5 + 2 - 8 - 3$$ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3 = \left( { - 7 + 12 + 9 - 2} \right)mn + 7 - 11$$ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3 = \left( {5 + 7} \right)mn - 4$$ - 7mn + 5,12mn + 2,9mn - 8, - 2mn - 3 = 12mn - 4$

Therefore, the obtained sum of the given terms are $12mn - 4$.

(iv) $a + b - 3,b - a + 3,a - b + 3$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $a$ and the variable $b$ then add the coefficients.

$a + b - 3,b - a + 3,a - b + 3 = a + b - 3 + b - a + 3 + a - b + 3$$a + b - 3,b - a + 3,a - b + 3 = \left( {a - a + a} \right) + \left( {b + b - b} \right) - 3 + 3 + 3$$a + b - 3,b - a + 3,a - b + 3 = a + b + 3$

Therefore, the obtained sum of the given terms are $a + b + 3$.

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

Ans: Combine the like terms (the terms which have the same variables and power) of variable $x$, the variable $y$ and the variable $xy$ then add the coefficients.

$14x + 10y - 12xy - 13,18 - 7x - 10y + 8xy,4xy = 14x + 10y - 12xy - 13 + 18 - 7x - 10y + 8xy + 4xy$$14x + 10y - 12xy - 13,18 - 7x - 10y + 8xy,4xy = 14x - 7x + 10y - 10y - 12xy + 8xy + 4xy - 13 + 18$$14x + 10y - 12xy - 13,18 - 7x - 10y + 8xy,4xy = 7x + 0y + 0xy + 5$$14x + 10y - 12xy - 13,18 - 7x - 10y + 8xy,4xy = 7x + 5$

Therefore, the obtained sum of the given terms are $7x + 5$.

(vi) $5m - 7n,3n - 4m + 2,2m - 3mn - 5$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $m$ and the variable $n$ then add the coefficients.

$5m - 7n,3n - 4m + 2,2m - 3mn - 5 = 5m - 7n + 3n - 4m + 2 + 2m - 3mn - 5$$5m - 7n,3n - 4m + 2,2m - 3mn - 5 = 5m - 4m + 2m - 7n + 3n - 3mn + 2 - 5$$5m - 7n,3n - 4m + 2,2m - 3mn - 5 = \left( {5 - 4 + 2} \right)m + \left( { - 7 + 3} \right)n - 3mn - 3$$5m - 7n,3n - 4m + 2,2m - 3mn - 5 = 3m - 4n - 3mn - 3$

Therefore, the obtained sum of the given terms are $3m - 4n - 3mn - 3$.

(vii) $4{x^2}y, - 3x{y^2}, - 5x{y^2},5{x^2}y$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${x^2}y$ and the variable $x{y^2}$ then add the coefficients.

$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$$4{x^2}y, - 3x{y^2}, - 5x{y^2},5{x^2}y = 4{x^2}y + 5{x^2}y + \left( { - 3x{y^2}} \right) + \left( { - 5x{y^2}} \right)$$4{x^2}y, - 3x{y^2}, - 5x{y^2},5{x^2}y = 9{x^2}y - 8x{y^2}$

Therefore, the obtained sum of the given terms are $9{x^2}y - 8x{y^2}$.

(viii) $3{p^2}{q^2} - 4pq + 5, - 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2}$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${p^2}{q^2}$ and the variable $pq$ then add the coefficients.

$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 + \left( { - 10{p^2}{q^2}} \right) + 15 + 9pq + 7{p^2}{q^2}$$3{p^2}{q^2} - 4pq + 5, - 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2} = 3{p^2}{q^2} - 10{p^2}{q^2} + 7{p^2}{q^2} - 4pq + 9pq + 5 + 15$$3{p^2}{q^2} - 4pq + 5, - 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2} = \left( {3 - 10 + 7} \right){p^2}{q^2} + \left( { - 4 + 9} \right)pq + 20$$3{p^2}{q^2} - 4pq + 5, - 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2} = 0{p^2}{q^2} + 5pq + 20$$3{p^2}{q^2} - 4pq + 5, - 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2} = 5pq + 20$

Therefore, the obtained sum of the given terms are $5pq + 20$.

(ix) $ab - 4a,4b - ab,4a - 4b$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $a$, the variable $b$ and the variable $ab$ then add the coefficients.

$ab - 4a,4b - ab,4a - 4b = ab - 4a + 4b - ab + 4a - 4b$$ab - 4a,4b - ab,4a - 4b = - 4a + 4a + 4b - 4b + ab - ab$

$ab - 4a,4b - ab,4a - 4b = 0 + 0 + 0$

$ab - 4a,4b - ab,4a - 4b = 0$

Therefore, the obtained sum of the given terms is $0$.

(x) ${x^2} - {y^2} - 1,{y^2} - 1 - {x^2},1 - {x^2} - {y^2}$

Ans: Combine the like terms (terms which have the same variables and power) of variable ${x^2}$ and the variable ${y^2}$ then add the coefficients.

${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}$${x^2} - {y^2} - 1,{y^2} - 1 - {x^2},1 - {x^2} - {y^2} = {x^2} - {x^2} - {x^2} - {y^2} + {y^2} - {y^2} - 1 - 1 + 1$${x^2} - {y^2} - 1,{y^2} - 1 - {x^2},1 - {x^2} - {y^2} = \left( {1 - 1 - 1} \right){x^2} + \left( { - 1 + 1 - 1} \right){y^2} - 1 - 1 + 1$${x^2} - {y^2} - 1,{y^2} - 1 - {x^2},1 - {x^2} - {y^2} = - {x^2} - {y^2} - 1$

Therefore, the obtained sum of the given terms are $ - {x^2} - {y^2} - 1$.

3. Subtract:

(i) $ - 5{y^2}$ from ${y^2}$

Ans: Subtract the coefficients of ${y^2}$.

${y^2} - \left( { - 5{y^2}} \right) = {y^2} + 5{y^2}$

${y^2} - \left( { - 5{y^2}} \right) = 6{y^2}$

Therefore, the obtained term is $6{y^2}$.

(ii) $6xy$ from $ - 12xy$

Ans: Subtract the coefficients of $xy$.

$ - 12xy - \left( {6xy} \right) = - 12xy - 6xy$

$ - 12xy - \left( {6xy} \right) = - 18xy$

Therefore, the obtained term is $ - 18xy$.

(iii) $\left( {a - b} \right)$ from $\left( {a + b} \right)$

Ans: Combine the like terms (the terms which have the same variables and power) of variable $a$ and the variable $b$ then subtract the coefficients.

$\left( {a + b} \right) - \left( {a - b} \right) = a + b - a + b$

$\left( {a + b} \right) - \left( {a - b} \right) = a - a + b + b$

$\left( {a + b} \right) - \left( {a - b} \right) = 2b$

Therefore, the obtained term is $2b$.

(iv) $a\left( {b - 5} \right)$ from $b\left( {5 - a} \right)$

Ans: Solve the brackets then combine the like terms (the terms which have the same variables and power) of variable $ab$ and then subtract the coefficients.

$b\left( {5 - a} \right) - a\left( {b - 5} \right) = 5b - ab - ab + 5a$

$b\left( {5 - a} \right) - a\left( {b - 5} \right) = 5b - 2ab + 5a$

$b\left( {5 - a} \right) - a\left( {b - 5} \right) = 5a + 5b - 2ab$

Therefore, the obtained term is $5a + 5b - 2ab$.

(v) $ - {m^2} + 5mn$ from $4{m^2} - 3mn + 8$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${m^2}$ and the variable $mn$ then subtract the coefficients.

$4{m^2} - 3mn + 8 - \left( { - {m^2} + 5mn} \right) = 4{m^2} - 3mn + 8 + {m^2} - 5mn$$4{m^2} - 3mn + 8 - \left( { - {m^2} + 5mn} \right) = 4{m^2} + {m^2} - 3mn - 5mn + 8$$4{m^2} - 3mn + 8 - \left( { - {m^2} + 5mn} \right) = 5{m^2} - 8mn + 8$

Therefore, the obtained term is $5{m^2} - 8mn + 8$.

(vi) $ - {x^2} + 10x - 5$ from $5x - 10$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${x^2}$ and the variable $x$ then subtract the coefficients.

$5x - 10 - \left( { - {x^2} + 10x - 5} \right) = 5x - 10 + {x^2} - 10x + 5$$5x - 10 - \left( { - {x^2} + 10x - 5} \right) = {x^2} + 5x - 10x - 10 + 5$

$5x - 10 - \left( { - {x^2} + 10x - 5} \right) = {x^2} - 5x - 5$

Therefore, the obtained term is ${x^2} - 5x - 5$.

(vii) $5{a^2} - 7ab + 5{b^2}$ from $3ab - 2{a^2} - 2{b^2}$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${a^2}$, the variable ${b^2}$ and the variable $ab$ then subtract the coefficients.

$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}$$3ab - 2{a^2} - 2{b^2} - \left( {5{a^2} - 7ab + 5{b^2}} \right) = 3ab + 7ab - 2{a^2} - 5{a^2} - 2{b^2} - 5{b^2}$$3ab - 2{a^2} - 2{b^2} - \left( {5{a^2} - 7ab + 5{b^2}} \right) = 10ab - 7{a^2} - 7{b^2}$$3ab - 2{a^2} - 2{b^2} - \left( {5{a^2} - 7ab + 5{b^2}} \right) = - 7{a^2} - 7{b^2} + 10ab$

Therefore, the obtained term is $ - 7{a^2} - 7{b^2} + 10ab$.

(viii) $4pq - 5{q^2} - 3{p^2}$ from $5{p^2} + 3{q^2} - pq$

Ans: Combine the like terms (the terms which have the same variables and power) of variable ${p^2}$, the variable ${q^2}$ and the variable $pq$ then subtract the coefficients.

$5{p^2} + 3{q^2} - pq - \left( {4pq - 5{q^2} - 3{p^2}} \right) = 5{p^2} + 3{q^2} - pq - 4pq + 5{q^2} + 3{p^2}$$5{p^2} + 3{q^2} - pq - \left( {4pq - 5{q^2} - 3{p^2}} \right) = 5{p^2} + 3{p^2} + 3{q^2} + 5{q^2} - pq - 4pq$$5{p^2} + 3{q^2} - pq - \left( {4pq - 5{q^2} - 3{p^2}} \right) = 8{p^2} + 8{q^2} - 5pq$

Therefore, the obtained term is $8{p^2} + 8{q^2} - 5pq$.

4.

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

Ans: Suppose, the expression is $a$ which should be added to ${x^2} + xy + {y^2}$ to obtain $2{x^2} + 3xy$. Hence, the equation is ${x^2} + xy + {y^2} + a = 2{x^2} + 3xy$.

Combine the like terms (the terms which have the same variables and power) of variable ${x^2}$ and $xy$ then simplify.

${x^2} + xy + {y^2} + a = 2{x^2} + 3xy$

$a = 2{x^2} + 3xy - \left( {{x^2} + xy + {y^2}} \right)$

$a = 2{x^2} + 3xy - {x^2} - xy - {y^2}$

$a = 2{x^2} - {x^2} - {y^2} + 3xy - xy$

$a = {x^2} - {y^2} + 2xy$

Therefore, ${x^2} - {y^2} + 2xy$ should be added.

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

Ans: Suppose, the expression is $x$ which should be subtracted from $2a + 8b + 10$ to obtain $ - 3a + 7b + 16$. Hence, the equation is $2a + 8b + 10 - x = - 3a + 7b + 16$.

Subtract the term $2a + 8b + 10$ from $ - 3a + 7b + 16$ then combine the like terms (the terms which have the same variables and power) of variable $a$ and $b$ then simplify.

$2a + 8b + 10 - x = - 3a + 7b + 16$

$ - x = - 3a + 7b + 16 - \left( {2a + 8b + 10} \right)$

$ - x = - 3a + 7b + 16 - 2a - 8b - 10$

$ - x = - 3a - 2a + 7b - 8b + 16 - 10$

$ - x = - 5a - b + 6$

$x = 5a + b - 6$

Therefore, $5a + b - 6$ should be subtracted.

5. What should be taken away from $3{x^2} - 4{y^2} + 5xy + 20$ to obtain $ - {x^2} - {y^2} + 6xy + 20$ ?

Ans: Suppose that $p$ is subtracted from $3{x^2} - 4{y^2} + 5xy + 20$ to get the result as $ - {x^2} - {y^2} + 6xy + 20$. Hence, the equation is $3{x^2} - 4{y^2} + 5xy + 20 - p = - {x^2} - {y^2} + 6xy + 20$.

Now subtract $3{x^2} - 4{y^2} + 5xy + 20$ from $ - {x^2} - {y^2} + 6xy + 20$. Combine the like terms (the terms which have the same variables and power) of variable ${x^2}$, the variable ${y^2}$, and the variable $xy$ then subtract the coefficients.

$3{x^2} - 4{y^2} + 5xy + 20 - p = - {x^2} - {y^2} + 6xy + 20$$ - p = - {x^2} - {y^2} + 6xy + 20 - \left( {3{x^2} - 4{y^2} + 5xy + 20} \right)$$ - p = - {x^2} - {y^2} + 6xy + 20 - 3{x^2} + 4{y^2} - 5xy - 20$$ - p = - {x^2} - 3{x^2} - {y^2} + 4{y^2} + 6xy - 5xy + 20 - 20$

$ - p = - 4{x^2} + 3{y^2} + xy + 0$

$p = 4{x^2} - 3{y^2} - xy$

Hence, $4{x^2} - 3{y^2} - xy$ should be taken away from $3{x^2} - 4{y^2} + 5xy + 20$.

6.

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

Ans: The equation according to the question is as follows.

$\left( {3x - y + 11} \right) + \left( { - y - 11} \right) - \left( {3x - y - 11} \right)$

Now, simplify the brackets and combine the like terms (the terms which have the same variables and power) of the variable $x$, the variable $y$ and the constants to simplify.

$\left( {3x - y + 11} \right) + \left( { - y - 11} \right) - \left( {3x - y - 11} \right) = 3x - y + 11 - y - 11 - 3x + y + 11$$\left( {3x - y + 11} \right) + \left( { - y - 11} \right) - \left( {3x - y - 11} \right) = 3x - 3x - y - y + y + 11 - 11 + 11$$\left( {3x - y + 11} \right) + \left( { - y - 11} \right) - \left( {3x - y - 11} \right) = 0x + \left( { - 1 - 1 + 1} \right)y + 11$$\left( {3x - y + 11} \right) + \left( { - y - 11} \right) - \left( {3x - y - 11} \right) = - y + 11$

Therefore, the required expression is $ - y + 11$.

(b) From the sum of $4 + 3x$ and $5 - 4x + 2{x^2}$ , subtract the sum of $3{x^2} - 5x$ and $ - {x^2} + 2x + 5$.

Ans: The equation according to the question is as follows.

$\left[ {\left( {4 + 3x} \right) + \left( {5 - 4x + 2{x^2}} \right)} \right] - \left[ {\left( {3{x^2} - 5x} \right) + \left( { - {x^2} + 2x + 5} \right)} \right]$

Now, simplify the brackets and add the terms.

$\left[ \left( 4+3x \right)+\left( 5-4x+2{{x}^{2}} \right) \right]-\left[ \left( 3{{x}^{2}}-5x \right)+\left( -{{x}^{2}}+2x+5 \right) \right]=$

$\left[ 4+3x+5-4x+2{{x}^{2}} \right]-\left[ 3{{x}^{2}}-5x-{{x}^{2}}+2x+5 \right]$

$\Rightarrow \left[ \left( 4+3x \right)+\left( 5-4x+2{{x}^{2}} \right) \right]-\left[ \left( 3{{x}^{2}}-5x \right)+\left( -{{x}^{2}}+2x+5 \right) \right]$

$=\left[ 2{{x}^{2}}+3x-4x+4+5 \right]-\left[ 3{{x}^{2}}-{{x}^{2}}-5x+2x+5 \right]$

$\Rightarrow \left[ \left( 4+3x \right)+\left( 5-4x+2{{x}^{2}} \right) \right]-\left[ \left( 3{{x}^{2}}-5x \right)+\left( -{{x}^{2}}+2x+5 \right) \right]$

$=\left[ 2{{x}^{2}}-x+9 \right]-\left[ 2{{x}^{2}}-3x+5 \right]$

Then solve the bigger brackets, combine the like terms (the terms which have the same variables and power) of the variable $x$, the variable ${x^2}$ and the constants to simplify.

$\left[ {\left( {4 + 3x} \right) + \left( {5 - 4x + 2{x^2}} \right)} \right] - \left[ {\left( {3{x^2} - 5x} \right) + \left( { - {x^2} + 2x + 5} \right)} \right] = 2{x^2} - x + 9 - 2{x^2} + 3x - 5$$\left[ {\left( {4 + 3x} \right) + \left( {5 - 4x + 2{x^2}} \right)} \right] - \left[ {\left( {3{x^2} - 5x} \right) + \left( { - {x^2} + 2x + 5} \right)} \right] = 2{x^2} - 2{x^2} - x + 3x + 9 - 5$$\left[ {\left( {4 + 3x} \right) + \left( {5 - 4x + 2{x^2}} \right)} \right] - \left[ {\left( {3{x^2} - 5x} \right) + \left( { - {x^2} + 2x + 5} \right)} \right] = 2x + 4$

Therefore, the required solution is $2x + 4$.

## NCERT Solutions For Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2

Opting for the NCERT solutions for Ex 12.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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