NCERT Solutions for Class 7 Maths Exercise 12.2

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NCERT Solutions for Class 7 Maths Algebraic Expressions

Class –VII Mathematics (Ex. 12.2)

Question 1.Simplify combining like terms:

1. {tex}21b – 32 + 7b – 20b{/tex}
2. {tex} – {z^2} + 13{z^2} – 5x + 7{z^3} – 15z{/tex}
3. {tex}p – \left( {p – q} \right) – q – \left( {q – p} \right){/tex}
4. {tex}3a – 2b – ab – \left( {a – b + ab} \right) + 3ab + b – a{/tex}
5. {tex}5{x^2}y – 5{x^2} + 3y{x^2} – 3{y^2} + {x^2} – {y^2} + 8x{y^2} – 3{y^2}{/tex}
6. {tex}\left( {3{y^2} + 5y – 4} \right) – \left( {8y – {y^2} – 4} \right){/tex}

Answer:

(i) {tex}21b – 32 + 7b – 20b = 21b + 7b – 20b – 32{/tex}

= {tex}28b – 20b – 32{/tex} = {tex}8b – 32{/tex}

(ii) {tex} – {z^2} + 13{z^2} – 5z + 7{z^3} – 15z = 7{z^3} + \left( { – {z^2} + 13{z^2}} \right) – \left( {5z + 15z} \right){/tex}

= {tex}7{z^3} + 12{z^2} – 20z{/tex}

(iii) {tex}p – \left( {p – q} \right) – q – \left( {q – p} \right) = p – p + q – q – q + p{/tex}

= {tex}p – p + p + q – q – q{/tex} = {tex}p – q{/tex}

(iv) {tex}3a – 2b – ab – \left( {a – b + ab} \right) + 3ab + b – a = 3a – 2b – ab – a + b – ab + 3ab + b – a{/tex}

= {tex}3a – a – a – 2b + b + b – ab – ab + 3ab{/tex}

= {tex}\left( {3a – a – a} \right) – \left( {2b – b – b} \right) – \left( {ab + ab – 3ab} \right){/tex}

= {tex}a – 0 – \left( { – ab} \right){/tex}

= {tex}a + ab{/tex}

(v) {tex}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}{/tex}

= {tex}\left( {5{x^2}y + 3{x^2}y} \right) + 8x{y^2} – \left( {5{x^2} – {x^2}} \right) – \left( {3{y^2} + {y^2} + 3{y^2}} \right){/tex}

= {tex}8{x^2}y + 8x{y^2} – 4{x^2} – 7{y^2}{/tex}

(vi) {tex}\left( {3{y^2} + 5y – 4} \right) – \left( {8y – {y^2} – 4} \right) = 3{y^2} + 5y – 4 – 8y + {y^2} + 4{/tex}

= {tex}\left( {3{y^2} + {y^2}} \right) + \left( {5y – 8y} \right) – \left( {4 – 4} \right){/tex}

= {tex}4{y^2} – 3y – 0{/tex} = {tex}4{y^2} – 3y{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.2

Question 2.Add:

1. {tex}3mn, – 5mn,8mn – 4mn{/tex}
2. {tex}t – 8tz,3tz – z,z – t{/tex}
3. {tex} – 7mn + 5,12mn + 2,9mn – 8, – 2mn – 3{/tex}
4. {tex}a + b – 3,b – a + 3,a – b + 3{/tex}
5. {tex}14x + 10y – 12xy – 13,18 – 7x – 10y + 8xy,4xy{/tex}
6. {tex}5m – 7n,3n – 4m + 2,2m – 3mn – 5{/tex}
7. {tex}4{x^2}y, – 3x{y^2}, – 5x{y^2},5{x^2}y{/tex}
8. {tex}3{p^2}{q^2} – 4pq + 5, – 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2}{/tex}
9. {tex}ab – 4a,4b – ab,4a – 4b{/tex}
10. {tex}{x^2} – {y^2} – 1,{y^2} – 1 – {x^2},1 – {x^2} – {y^2}{/tex}

Answer:

(i) {tex}3mn, – 5mn,8mn, – 4mn = 3mn + \left( { – 5mn} \right) + 8mn + \left( { – 4mn} \right){/tex}

= {tex}\left( {3 – 5 + 8 – 4} \right)mn{/tex} = {tex}2mn{/tex}

(ii) {tex}t – 8tz,3tz – z,z – t = t – 8tz + 3tz – z + z – t{/tex}

= {tex}t – t – 8tz + 3tz – z + z{/tex}

= {tex}\left( {1 – 1} \right)t + \left( { – 8 + 3} \right)tz + \left( { – 1 + 1} \right)z{/tex}

= {tex}0 – 5tz + 0{/tex} = {tex} – 5tz{/tex}

(iii) {tex} – 7mn + 5,12mn + 2,9mn – 8, – 2mn – 3 = – 7mn + 5 + 12mn + 2 + 9mn – 8 + \left( { – 2mn} \right) – 3{/tex}

= {tex} – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3{/tex}

= {tex}\left( { – 7 + 12 + 9 – 2} \right)mn + 7 – 11{/tex}

= {tex}12mn – 4{/tex}

(iv) {tex}a + b – 3,b – a + 3,a – b + 3 = a + b – 3 + b – a + 3 + a – b + 3{/tex}

= {tex}\left( {a – a + a} \right) + \left( {b + b – b} \right) – 3 + 3 + 3{/tex}

= {tex}a + b + 3{/tex}

(v) {tex}14x + 10y – 12xy – 13,18 – 7x – 10y + 8xy,4xy = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy{/tex}

= {tex}14x – 7x + 10y – 10y – 12xy + 8xy + 4xy – 13 + 18{/tex}

= {tex}7x + 0y + 0xy + 5{/tex} = {tex}7x + 5{/tex}

(vi) {tex}5m – 7n,3n – 4m + 2,2m – 3mn – 5 = 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5{/tex}

= {tex}5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5{/tex}

= {tex}\left( {5 – 4 + 2} \right)m + \left( { – 7 + 3} \right)n – 3mn – 3{/tex}

= {tex}3m – 4n + 3mn – 3{/tex}

(vii) {tex}4{x^2}y, – 3x{y^2}, – 5x{y^2},5{x^2}y = 4{x^2}y + ( – 3x{y^2}) + ( – 5x{y^2}) + 5{x^2}y{/tex}

= {tex}4{x^2}y + 5{x^2}y – 3x{y^2} – 5x{y^2}{/tex}

= {tex}9{x^2}y – 8x{y^2}{/tex}

(viii) {tex}3{p^2}{q^2} – 4pq + 5, – 10{p^2}{q^2},15 + 9pq + 7{p^2}{q^2}{/tex} = {tex}3{p^2}{q^2} – 4pq + 5 + \left( { – 10{p^2}{q^2}} \right) + 15 + 9pq + 7{p^2}{q^2}{/tex}

= {tex}3{p^2}{q^2} – 10{p^2}{q^2} + 7{p^2}{q^2} + 4pq + 9pq + 5 + 15{/tex}

= {tex}\left( {3 – 10 + 7} \right){p^2}{q^2} + \left( { – 4 + 9} \right)pq + 20{/tex}

= {tex}0{p^2}{q^2} + 5pq + 20{/tex} = {tex}5pq + 20{/tex}

(ix) {tex}ab – 4a,4b – ab,4a – ab = ab – 4a + 4b – ab + 4a – ab{/tex}

= {tex} – 4a + 4a + 4b – 4b + ab – ab{/tex}

= {tex}0 + 0 + 0 = 0{/tex}

(x) {tex}{x^2} – {y^2} – 1,{y^2} – 1 – {x^2},1 – {x^2} – {y^2}{/tex} = {tex}{x^2} – {y^2} – 1 + {y^2} – 1 – {x^2} + 1 – {x^2} – {y^2}{/tex}

= {tex}{x^2} – {x^2} – {x^2} – {y^2} + {y^2} – {y^2} – 1 – 1 + 1{/tex}

= {tex}\left( {1 – 1 – 1} \right){x^2} + \left( { – 1 + 1 – 1} \right){y^2} – 1 – 1 + 1{/tex}

= {tex} – {x^2} – {y^2} – 1{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.2

Question 3.Subtract:

1. {tex} – 5{y^2}{/tex} from {tex}{y^2}{/tex}
2. {tex}6xy{/tex} from {tex} – 12xy{/tex}
3. {tex}\left( {a – b} \right){/tex} from {tex}\left( {a + b} \right){/tex}
4. {tex}a\left( {b – 5} \right){/tex} from {tex}b\left( {5 – a} \right){/tex}
5. {tex} – {m^2} + 5mn{/tex} from {tex}4{m^2} – 3mn + 8{/tex}
6. {tex} – {x^2} + 10x – 5{/tex} from {tex}5x – 10{/tex}
7. {tex}5{a^2} – 7ab + 5{b^2}{/tex} from {tex}3ab – 2{a^2} – 2{b^2}{/tex}
8. {tex}4pq – 5{q^2} – 3{p^2}{/tex} from {tex}5{p^2} + 3{q^2} – pq{/tex}

Answer:

(i) {tex}{y^2} – \left( { – 5{y^2}} \right){/tex} = {tex}{y^2} + 5{y^2}{/tex} = {tex}6{y^2}{/tex}

(ii) {tex} – 12xy – \left( {6xy} \right){/tex} = {tex} – 12xy – 6xy{/tex} = {tex} – 18xy{/tex}

(iii) {tex}\left( {a + b} \right) – \left( {a – b} \right){/tex} = {tex}a + b – a + b{/tex} = {tex}a – a + b + b{/tex} = {tex}2b{/tex}

(iv) {tex}b\left( {5 – a} \right) – a\left( {b – 5} \right){/tex} = {tex}5b – ab – ab + 5a{/tex} = {tex}5b – 2ab + 5a{/tex} = {tex}5a + 5b – 2ab{/tex}

(v) {tex}4{m^2} – 3mn + 8 – \left( { – {m^2} + 5mn} \right){/tex} = {tex}4{m^2} – 3mn + 8 + {m^2} – 5mn{/tex}

= {tex}4{m^2} + {m^2} – 3mn – 5mn + 8{/tex}

= {tex}5{m^2} – 8mn + 8{/tex}

(vi) {tex}5x – 10 – \left( { – {x^2} + 10x – 5} \right){/tex} = {tex}5x – 10 + {x^2} – 10x + 5{/tex}

= {tex}{x^2} + 5x – 10x – 10 + 5{/tex} = {tex}{x^2} – 5x – 5{/tex}

(vii) {tex}3ab – 2{a^2} – 2{b^2} – \left( {5{a^2} – 7ab + 5{b^2}} \right){/tex} = {tex}3ab – 2{a^2} – 2{b^2} – 5{a^2} + 7ab – 5{b^2}{/tex}

= {tex}3ab + 7ab – 2{a^2} – 5{a^2} – 2{b^2} – 5{b^2}{/tex}

= {tex}10ab – 7{a^2} – 7{b^2}{/tex}

= {tex} – 7{a^2} – 7{b^2} + 10ab{/tex}

(viii) {tex}5{p^2} + 3{q^2} – pq – \left( {4pq – 5{q^2} – 3{p^2}} \right){/tex} = {tex}5{p^2} + 3{q^2} – pq – 4pq + 5{q^2} + 3{p^2}{/tex}

= {tex}5{p^2} + 3{p^2} + 3{q^2} + 5{q^2} – pq – 4pq{/tex}

= {tex}8{p^2} + 8{q^2} – 5pq{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.2

Question 4.(a) What should be added to {tex}{x^2} + xy + {y^2}{/tex} to obtain {tex}2{x^2} + 3xy?{/tex}

(b) What should be subtracted from {tex}2a + 8b + 10{/tex} to get {tex} – 3a + 7b + 16{/tex} ?

Answer:

(a) Let {tex}p{/tex} should be added.

Then according to question,

{tex}{x^2} + xy + {y^2} + p = 2{x^2} + 3xy{/tex} {tex} \Rightarrow {/tex} {tex}p = 2{x^2} + 3xy – \left( {{x^2} + xy + {y^2}} \right){/tex}

{tex} \Rightarrow {/tex} {tex}p = 2{x^2} + 3xy – {x^2} – xy – {y^2}{/tex} {tex} \Rightarrow {/tex} {tex}p = 2{x^2} – {x^2} – {y^2} + 3xy – xy{/tex}

{tex} \Rightarrow {/tex} {tex}p = {x^2} – {y^2} + 2xy{/tex}

Hence, {tex}{x^2} – {y^2} + 2xy{/tex} should be added.

(b) Let {tex}q{/tex} should be subtracted.

Then according to question,

{tex}2a + 8b + 10 – q = – 3a + 7b + 16{/tex} {tex} \Rightarrow {/tex} {tex} – q = – 3a + 7b + 16 – \left( {2a + 8b + 10} \right){/tex}

{tex} \Rightarrow {/tex} {tex} – q = – 3a + 7b + 16 – 2a – 8b – 10{/tex} {tex} \Rightarrow {/tex} {tex} – q = – 3a – 2a + 7b – 8b + 16 – 10{/tex}

{tex} \Rightarrow {/tex} {tex} – q = – 5a – b + 6{/tex} {tex} \Rightarrow {/tex}{tex}q = – \left( { – 5a – b + 6} \right){/tex}

{tex} \Rightarrow {/tex} {tex}q = 5a + b – 6{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.2

Question 5.What should be taken away from {tex}3{x^2} – 4{y^2} + 5xy + 20{/tex} to obtain {tex} – {x^2} – {y^2} + 6xy + 20{/tex} ?

Answer:

Let {tex}q{/tex} should be subtracted.

Then according to question,

{tex}3{x^2} – 4{y^2} + 5xy + 20 – q = – {x^2} – {y^2} + 6xy + 20{/tex}

{tex} \Rightarrow {/tex} {tex}q = 3{x^2} – 4{y^2} + 5xy + 20 – \left( { – {x^2} – {y^2} + 6xy + 20} \right){/tex}

{tex} \Rightarrow {/tex} {tex}q = 3{x^2} – 4{y^2} + 5xy + 20 + {x^2} + {y^2} – 6xy – 20{/tex}

{tex} \Rightarrow {/tex} {tex}q = 3{x^2} + {x^2} – 4{y^2} + {y^2} + 5xy – 6xy + 20 – 20{/tex}

{tex} \Rightarrow {/tex} {tex}q = 4{x^2} – 3{y^2} – xy + 0{/tex}

Hence, {tex}4{x^2} – 3{y^2} – xy{/tex} should be subtracted.

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NCERT Solutions for Class 7 Maths Exercise 12.2

Question 6.(a) From the sum of {tex}3x – y + 11{/tex} and {tex} – y – 11,{/tex} subtract the sum of {tex}3{x^2} – 5x{/tex} and {tex} – {x^2} + 2x + 5.{/tex}

Answer:

(a) According to question,

{tex}\left( {3x – y + 11} \right) + \left( { – y – 11} \right) – \left( {3x – y – 11} \right){/tex} = {tex}3x – y + 11 – y – 11 – 3x + y + 11{/tex}

= {tex}3x – 3x – y – y + y + 11 – 11 + 11{/tex}

= {tex}\left( {3 – 3} \right)x – \left( {1 + 1 – 1} \right)y + 11 + 11 – 11{/tex}

= {tex}0x – y + 11{/tex} = {tex} – y + 11{/tex}

(b) According to question,

{tex}\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]{/tex}

= {tex}\left[ {4 + 3x + 5 – 4x + 2{x^2}} \right] – \left[ {3{x^2} – 5x – {x^2} + 2x + 5} \right]{/tex}

= {tex}\left[ {2{x^2} + 3x – 4x + 5 + 4} \right] – \left[ {3{x^2} – {x^2} + 2x – 5x + 5} \right]{/tex}

= {tex}\left[ {2{x^2} – x + 9} \right] – \left[ {2{x^2} – 3x + 5} \right]{/tex}

= {tex}2{x^2} – x + 9 – 2{x^2} + 3x – 5{/tex}

= {tex}2{x^2} – 2{x^2} – x + 3x + 9 – 5{/tex}

= {tex}2x + 4{/tex}

NCERT Solutions for Class 7 Maths Exercise 12.2

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