Simple Equations Ex-4.3

NCERT solutions for Maths Simple Equations 

NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.3)

Question 1.Solve the following equations:

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

(b) $5t + 28 = 10$

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

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

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

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

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

(h) $6z + 10 = – 2$

(i) $\frac{{3l}}{2} = \frac{2}{3}$

(j) $\frac{{2b}}{3} – 5 = 3$

Answer:

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

$\Rightarrow$ $2y = \frac{{32}}{2}$ $\Rightarrow$ $2y = 16$ $\Rightarrow$ $y = \frac{{16}}{2}$

$\Rightarrow$ $y = 8$

(b) $5t + 28 = 10$ $\Rightarrow$ $5t = 10 – 28$ $\Rightarrow$ $5t = – 18$

$\Rightarrow$ $t = \frac{{ – 18}}{5}$

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

$\Rightarrow$ $a = – 1 \times 5$ $\Rightarrow$ $a = – 5$

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

$\Rightarrow$ $q = – 2 \times 4$ $\Rightarrow$ $q = – 8$

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

$\Rightarrow$ $x = \frac{{20}}{5}$ $\Rightarrow$ $x = 4$

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

$\Rightarrow$ $x = \frac{{25}}{{2 \times 5}}$ $\Rightarrow$ $x = \frac{5}{2}$

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

$\Rightarrow$ $7m = \frac{7}{2}$ $\Rightarrow$ $m = \frac{7}{{2 \times 7}}$ $\Rightarrow$ $m = \frac{1}{2}$

(h) $6z + 10 = – 2$ $\Rightarrow$ $6z = – 2 – 10$ $\Rightarrow$ $6z = – 12$

$\Rightarrow$ $z = \frac{{ – 12}}{6}$ $\Rightarrow$ $z = – 2$

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

$\Rightarrow$ $l = \frac{4}{{3 \times 3}}$ $\Rightarrow$$l = \frac{4}{9}$

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

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

$\Rightarrow$ $b = 12$

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 2.Solve the following equations:

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

(b) $3\left( {n – 5} \right) = 21$

(c) $3\left( {n – 5} \right) = – 21$

(d) $3 – 2\left( {2 – y} \right) = 7$

(e) $– 4\left( {2 – x} \right) = 9$

(f) $4\left( {2 – x} \right) = 9$

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

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

Answer:

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

$\Rightarrow$ $x=6-4$ $\Rightarrow$ $x=2$

(b)$\Rightarrow$ $n=7+5$ $\Rightarrow$ $n=12$

(c)$\Rightarrow$ $n=-7+5$ $\Rightarrow$ $n=-2$

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

$\Rightarrow$ $-y=-4$ $\Rightarrow$ $y=4$

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

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

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

$\Rightarrow$ $p=7$

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

$\Rightarrow$ $p=7$

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 3.Solve the following equations:

(a) $4 = 5\left( {p – 2} \right)$

(b) $– 4 = 5\left( {p – 2} \right)$

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

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

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

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

Answer:

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

$\Rightarrow$ $x = 6 – 4$ $\Rightarrow$ $x = 2$

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

$\Rightarrow$ $n = 7 + 5$ $\Rightarrow$ $n = 12$

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

$\Rightarrow$ $n = – 7 + 5$ $\Rightarrow$ $n = – 2$

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

$\Rightarrow$ $2 – y = \frac{4}{{ – 2}}$ $\Rightarrow$ $2 – y = – 2$ $\Rightarrow$ $– y = – 2 – 2$

$\Rightarrow$ $– y = – 4$ $\Rightarrow$ $y = 4$

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

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

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

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

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

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

$\Rightarrow$ $p = 7$

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

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

$\Rightarrow$ $p = 7$

NCERT Solutions for Class 7 Maths Exercise 4.3

Question 4.

(a) Construct 3 equations starting with $x = 2.$

(b) Construct 3 equations starting with $x = – 2.$

Answer:

(a) 3 equations starting with $x = 2.$

(i) $x = 2$

Multiplying both sides by 10, $10x = 20$

Adding 2 both sides $10x + 2 = 20 + 2$ = $10x + 2 = 22$

(ii) $x = 2$

Multiplying both sides by 5 $5x = 10$

Subtracting 3 from both sides $5x – 3 = 10 – 3$ = $5x – 3 = 7$

(iii) $x = 2$

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

(b) 3 equations starting with $x = – 2.$

(i) $x = – 2$

Multiplying both sides by 3 $3x = – 6$

(ii) $x = – 2$

Multiplying both sides by 3 $3x = – 6$

Adding 7 to both sides $3x + 7 = – 6 + 7$ = $3x + 7 = 1$

(iii) $x = – 2$

Multiplying both sides by 3 $3x = – 6$

Adding 10 to both sides $3x + 10 = – 6 + 10$ = $3x + 10 = 4$