NCERT Solutions for Class 7 Maths Exercise 4.3

NCERT Solutions for Class 7 Maths Exercise 4.3 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Simple Equations Download as PDF

NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.3)

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}

Answer:

(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}

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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}

Answer:

(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}

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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}

Answer:

(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}

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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}

Answer:

(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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