# Solve the following equations:$(a).\ 10p=100$$(b).\ 10p+10=100$$(c).\ \frac{p}{4}=5$$(d).\ \frac{-p}{3}=5$$(e).\ \frac{3p}{4}=6$$(f).\ 3s=-9$$(g).\ 3s+12=0$$(h).\ 3s=0$$(i).\ 2q=6$$(j).\ 2q-6=0$$(k).\ 2q+6=0$$(l).\ 2q+6=12$

To do:

We have to solve the given equations.

Solution:

(a) $10p=100$

Dividing both the sides by 10 we get,

$\frac{10p}{10}=\frac{100}{10}$

$p=10$

(b) $10p+10 =100$

Subtracting 10 from both sides we get,

$10p+10-10=100-10$

$10p=90$

Dividing both the sides by 10 we get,

$\frac{10p}{10}=\frac{90}{10}$

$p=9$

(c) $\frac{p}{4}=5$

Multiplying both the sides by 4 we get,

$(\frac{p}{4})\times 4=5\times 4$

$p=20$

(d) $\frac{-p}{3}=5$

Multiplying both the sides by $-3$, we get,

$(\frac{-p}{3})\times (-3)=5\times (-3)$

$p =-15$

(e) $\frac{3p}{4}=6$

Multiplying both the sides by 4, we get,

$(\frac{3p}{4})\times 4=6\times 4$

$3p=24$

Dividing both the sides by 3 we get,

$\frac{3p}{3}=\frac{24}{3}$

$p=8$

(f) $3s=-9$

Dividing both the sides by 3,

$\frac{3s}{3}=\frac{-9}{3}$

$s=-3$

(g) $3s+12=0$

Subtracting 12 from both the sides of the equation we get,

$3s+12- 12=0- 12$

$3s =-12$

Dividing both the sides by 3 we get,

$\frac{3s}{3}=\frac{-12}{3}$

$s=-4$

(h) $3s=0$

Dividing both the sides by 3 we get,

$\frac{3s}{3}=\frac{0}{3}$

$s=0$

(i) $2q=6$

Dividing both the sides by 2 we get,

$\frac{2q}{2}=\frac{6}{2}$

$q=3$

(j) $2q- 6=0$

Adding 6 to both sides of the equation we get,

$2q- 6+6=0+6$

$2q=6$

Dividing both the sides by 2 we get,

$\frac{2q}{2}=\frac{6}{2}$

$q=3$

(k) $2q+6=0$

Subtracting 6 from both the sides of the equation we get,

$2q+6-6=0- 6$

$2q =- 6$

Dividing both the sides by 2 we get,

$\frac{2q}{2}=-\frac{6}{2}$

$q=-3$

(l) $2q+6 =12$

Subtracting 6 from both the sides of the equation we get,

$2q+6-6=12-6$

$2q=6$

Dividing both the sides by 2 we get,

$\frac{2q}{2}=\frac{6}{2}$

$q=3$

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Updated on: 10-Oct-2022

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