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NCERT Solutions for Class 7 Maths Exercise 4.3

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NCERT solutions for Maths Simple Equations

NCERT Solutions for Class 7 Maths Simple Equations

Question 1.Solve the following equations:

(a) {tex}2y + \frac{5}{2} = \frac{{37}}{2}{/tex}

(b) {tex}5t + 28 = 10{/tex}

(c) {tex}\frac{a}{5} + 3 = 2{/tex}

(d) {tex}\frac{q}{4} + 7 = 5{/tex}

(e) {tex}\frac{5}{2}x = 10{/tex}

(f) {tex}\frac{5}{2}x = \frac{{25}}{4}{/tex}

(g) {tex}7m + \frac{{19}}{2} = 13{/tex}

(h) {tex}6z + 10 = – 2{/tex}

(i) {tex}\frac{{3l}}{2} = \frac{2}{3}{/tex}

(j) {tex}\frac{{2b}}{3} – 5 = 3{/tex}

(a) {tex}2y + \frac{5}{2} = \frac{{37}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}2y = \frac{{37}}{2} – \frac{5}{2}{/tex} {tex} \Rightarrow {/tex} {tex}2y = \frac{{37 – 5}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}2y = \frac{{32}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}2y = 16{/tex} {tex} \Rightarrow {/tex} {tex}y = \frac{{16}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}y = 8{/tex}

(b) {tex}5t + 28 = 10{/tex} {tex} \Rightarrow {/tex} {tex}5t = 10 – 28{/tex} {tex} \Rightarrow {/tex} {tex}5t = – 18{/tex}

{tex} \Rightarrow {/tex} {tex}t = \frac{{ – 18}}{5}{/tex}

(c) {tex}\frac{a}{5} + 3 = 2{/tex} {tex} \Rightarrow {/tex} {tex}\frac{a}{5} = 2 – 3{/tex} {tex} \Rightarrow {/tex} {tex}\frac{a}{5} = – 1{/tex}

{tex} \Rightarrow {/tex} {tex}a = – 1 \times 5{/tex} {tex} \Rightarrow {/tex} {tex}a = – 5{/tex}

(d) {tex}\frac{q}{4} + 7 = 5{/tex} {tex} \Rightarrow {/tex} {tex}\frac{q}{4} = 5 – 7{/tex} {tex} \Rightarrow {/tex} {tex}\frac{q}{4} = – 2{/tex}

{tex} \Rightarrow {/tex} {tex}q = – 2 \times 4{/tex} {tex} \Rightarrow {/tex} {tex}q = – 8{/tex}

(e) {tex}\frac{5}{2}x = 10{/tex} {tex} \Rightarrow {/tex} {tex}5x = 10 \times 2{/tex} {tex} \Rightarrow {/tex} {tex}5x = 20{/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{20}}{5}{/tex} {tex} \Rightarrow {/tex} {tex}x = 4{/tex}

(f) {tex}\frac{5}{2}x = \frac{{25}}{4}{/tex} {tex} \Rightarrow {/tex} {tex}5x = \frac{{25}}{4} \times 2{/tex} {tex} \Rightarrow {/tex} {tex}5x = \frac{{25}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}x = \frac{{25}}{{2 \times 5}}{/tex} {tex} \Rightarrow {/tex} {tex}x = \frac{5}{2}{/tex}

(g) {tex}7m + \frac{{19}}{2} = 13{/tex} {tex} \Rightarrow {/tex} {tex}7m = 13 – \frac{{19}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}7m = \frac{{26 – 19}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}7m = \frac{7}{2}{/tex} {tex} \Rightarrow {/tex} {tex}m = \frac{7}{{2 \times 7}}{/tex} {tex} \Rightarrow {/tex} {tex}m = \frac{1}{2}{/tex}

(h) {tex}6z + 10 = – 2{/tex} {tex} \Rightarrow {/tex} {tex}6z = – 2 – 10{/tex} {tex} \Rightarrow {/tex} {tex}6z = – 12{/tex}

{tex} \Rightarrow {/tex} {tex}z = \frac{{ – 12}}{6}{/tex} {tex} \Rightarrow {/tex} {tex}z = – 2{/tex}

(i) {tex}\frac{{3l}}{2} = \frac{2}{3}{/tex} {tex} \Rightarrow {/tex} {tex}3l = \frac{2}{3} \times 2{/tex} {tex} \Rightarrow {/tex} {tex}3l = \frac{4}{3}{/tex}

{tex} \Rightarrow {/tex} {tex}l = \frac{4}{{3 \times 3}}{/tex} {tex} \Rightarrow {/tex}{tex}l = \frac{4}{9}{/tex}

(j) {tex}\frac{{2b}}{3} – 5 = 3{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2b}}{3} = 3 + 5{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{2b}}{3} = 8{/tex}

{tex} \Rightarrow {/tex} {tex}2b = 8 \times 3{/tex} {tex} \Rightarrow {/tex} {tex}2b = 24{/tex} {tex} \Rightarrow {/tex} {tex}b = \frac{{24}}{2}{/tex}

{tex} \Rightarrow {/tex} {tex}b = 12{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 2.Solve the following equations:

(a) {tex}2\left( {x + 4} \right) = 12{/tex}

(b) {tex}3\left( {n – 5} \right) = 21{/tex}

(c) {tex}3\left( {n – 5} \right) = – 21{/tex}

(d) {tex}3 – 2\left( {2 – y} \right) = 7{/tex}

(e) {tex} – 4\left( {2 – x} \right) = 9{/tex}

(f) {tex}4\left( {2 – x} \right) = 9{/tex}

(g) {tex}4 + 5\left( {p – 1} \right) = 34{/tex}

(h) {tex}34 – 5\left( {p – 1} \right) = 4{/tex}

(a) {tex}2\left( x+4 \right)=12{/tex} {tex}\Rightarrow {/tex} {tex}x+4=\frac{12}{2}{/tex} {tex}\Rightarrow {/tex} {tex}x+4=6{/tex}

{tex}\Rightarrow {/tex} {tex}x=6-4{/tex} {tex}\Rightarrow {/tex} {tex}x=2{/tex}

(b){tex}\Rightarrow {/tex} {tex}n=7+5{/tex} {tex}\Rightarrow {/tex} {tex}n=12{/tex}

(c){tex}\Rightarrow {/tex} {tex}n=-7+5{/tex} {tex}\Rightarrow {/tex} {tex}n=-2{/tex}

(d){tex}\Rightarrow {/tex} {tex}2-y=\frac{4}{-2}{/tex} {tex}\Rightarrow {/tex} {tex}2-y=-2{/tex} {tex}\Rightarrow {/tex} {tex}-y=-2-2{/tex}

{tex}\Rightarrow {/tex} {tex}-y=-4{/tex} {tex}\Rightarrow {/tex} {tex}y=4{/tex}

(e){tex}\Rightarrow {/tex} {tex}4x=9+8{/tex} {tex}\Rightarrow {/tex} {tex}4x=17{/tex} {tex}\Rightarrow {/tex} {tex}x=\frac{17}{4}{/tex}

(f){tex}\Rightarrow {/tex} {tex}-4x=9-8{/tex} {tex}\Rightarrow {/tex} {tex}-4x=1{/tex} {tex}\Rightarrow {/tex} {tex}x=\frac{-1}{4}{/tex}

(g){tex}\Rightarrow {/tex} {tex}p-1=\frac{30}{5}{/tex} {tex}\Rightarrow {/tex} {tex}p-1=6{/tex} {tex}\Rightarrow {/tex} {tex}p=6+1{/tex}

{tex}\Rightarrow {/tex} {tex}p=7{/tex}

(h){tex}\Rightarrow {/tex} {tex}p-1=\frac{-30}{-5}{/tex} {tex}\Rightarrow {/tex} {tex}p-1=6{/tex} {tex}\Rightarrow {/tex} {tex}p=6+1{/tex}

{tex}\Rightarrow {/tex} {tex}p=7{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 3.Solve the following equations:

(a) {tex}4 = 5\left( {p – 2} \right){/tex}

(b) {tex} – 4 = 5\left( {p – 2} \right){/tex}

(c) {tex} – 16 = – 5\left( {2 – p} \right){/tex}

(d) {tex}10 = 4 + 3\left( {t + 2} \right){/tex}

(e) {tex}28 = 4 + 3\left( {t + 5} \right){/tex}

(f) {tex}0 = 16 + 4\left( {m – 6} \right){/tex}

(a) {tex}2\left( {x + 4} \right) = 12{/tex} {tex} \Rightarrow {/tex} {tex}x + 4 = \frac{{12}}{2}{/tex} {tex} \Rightarrow {/tex} {tex}x + 4 = 6{/tex}

{tex} \Rightarrow {/tex} {tex}x = 6 – 4{/tex} {tex} \Rightarrow {/tex} {tex}x = 2{/tex}

(b) {tex}3\left( {n – 5} \right) = 21{/tex} {tex} \Rightarrow {/tex} {tex}n – 5 = \frac{{21}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}n – 5 = 7{/tex}

{tex} \Rightarrow {/tex} {tex}n = 7 + 5{/tex} {tex} \Rightarrow {/tex} {tex}n = 12{/tex}

(c) {tex}3\left( {n – 5} \right) = – 21{/tex} {tex} \Rightarrow {/tex} {tex}n – 5 = \frac{{ – 21}}{3}{/tex} {tex} \Rightarrow {/tex} {tex}n – 5 = – 7{/tex}

{tex} \Rightarrow {/tex} {tex}n = – 7 + 5{/tex} {tex} \Rightarrow {/tex} {tex}n = – 2{/tex}

(d) {tex}3 – 2\left( {2 – y} \right) = 7{/tex} {tex} \Rightarrow {/tex} {tex} – 2\left( {2 – y} \right) = 7 – 3{/tex} {tex} \Rightarrow {/tex} {tex} – 2\left( {2 – y} \right) = 4{/tex}

{tex} \Rightarrow {/tex} {tex}2 – y = \frac{4}{{ – 2}}{/tex} {tex} \Rightarrow {/tex} {tex}2 – y = – 2{/tex} {tex} \Rightarrow {/tex} {tex} – y = – 2 – 2{/tex}

{tex} \Rightarrow {/tex} {tex} – y = – 4{/tex} {tex} \Rightarrow {/tex} {tex}y = 4{/tex}

(e) {tex} – 4\left( {2 – x} \right) = 9{/tex} {tex} \Rightarrow {/tex} {tex} – 4 \times 2 – x \times \left( { – 4} \right) = 9{/tex} {tex} \Rightarrow {/tex} {tex} – 8 + 4x = 9{/tex}

{tex} \Rightarrow {/tex} {tex}4x = 9 + 8{/tex} {tex} \Rightarrow {/tex} {tex}4x = 17{/tex} {tex} \Rightarrow {/tex} {tex}x = \frac{{17}}{4}{/tex}

(f) {tex}4\left( {2 – x} \right) = 9{/tex} {tex} \Rightarrow {/tex} {tex}4 \times 2 – x \times \left( 4 \right) = 9{/tex} {tex} \Rightarrow {/tex} {tex}8 – 4x = 9{/tex}

{tex} \Rightarrow {/tex} {tex} – 4x = 9 – 8{/tex} {tex} \Rightarrow {/tex} {tex} – 4x = 1{/tex} {tex} \Rightarrow {/tex} {tex}x = \frac{{ – 1}}{4}{/tex}

(g) {tex}4 + 5\left( {p – 1} \right) = 34{/tex} {tex} \Rightarrow {/tex} {tex}5\left( {p – 1} \right) = 34 – 4{/tex} {tex} \Rightarrow {/tex} {tex}5\left( {p – 1} \right) = 30{/tex}

{tex} \Rightarrow {/tex} {tex}p – 1 = \frac{{30}}{5}{/tex} {tex} \Rightarrow {/tex} {tex}p – 1 = 6{/tex} {tex} \Rightarrow {/tex} {tex}p = 6 + 1{/tex}

{tex} \Rightarrow {/tex} {tex}p = 7{/tex}

(h) {tex}34 – 5\left( {p – 1} \right) = 4{/tex} {tex} \Rightarrow {/tex} {tex} – 5\left( {p – 1} \right) = 4 – 34{/tex} {tex} \Rightarrow {/tex} {tex} – 5\left( {p – 1} \right) = – 30{/tex}

{tex} \Rightarrow {/tex} {tex}p – 1 = \frac{{ – 30}}{{ – 5}}{/tex} {tex} \Rightarrow {/tex} {tex}p – 1 = 6{/tex} {tex} \Rightarrow {/tex} {tex}p = 6 + 1{/tex}

{tex} \Rightarrow {/tex} {tex}p = 7{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 4.

(a) Construct 3 equations starting with {tex}x = 2.{/tex}

(b) Construct 3 equations starting with {tex}x = – 2.{/tex}

(a) 3 equations starting with {tex}x = 2.{/tex}

(i) {tex}x = 2{/tex}

Multiplying both sides by 10, {tex}10x = 20{/tex}

Adding 2 both sides {tex}10x + 2 = 20 + 2{/tex} = {tex}10x + 2 = 22{/tex}

(ii) {tex}x = 2{/tex}

Multiplying both sides by 5 {tex}5x = 10{/tex}

Subtracting 3 from both sides {tex}5x – 3 = 10 – 3{/tex} = {tex}5x – 3 = 7{/tex}

(iii) {tex}x = 2{/tex}

Dividing both sides by 5 {tex}\frac{x}{5} = \frac{2}{5}{/tex}

(b) 3 equations starting with {tex}x = – 2.{/tex}

(i) {tex}x = – 2{/tex}

Multiplying both sides by 3 {tex}3x = – 6{/tex}

(ii) {tex}x = – 2{/tex}

Multiplying both sides by 3 {tex}3x = – 6{/tex}

Adding 7 to both sides {tex}3x + 7 = – 6 + 7{/tex} = {tex}3x + 7 = 1{/tex}

(iii) {tex}x = – 2{/tex}

Multiplying both sides by 3 {tex}3x = – 6{/tex}

Adding 10 to both sides {tex}3x + 10 = – 6 + 10{/tex} = {tex}3x + 10 = 4{/tex}

NCERT Solutions for Class 7 Maths Exercise 4.3

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