Question

Solve the following question using a systematic method
1) 7x=28
2) 2x+7=13
3) -y+36=32
4) 2y-6=8

Answer 1

i) $7x=28$

Divide both sides by $7$

$\frac{7x}{7}=\frac{28}{7}$

$\Rightarrow x=4$

(ii) $2x+7=13$

Add $-7$ to both sides,

$2x+7-7=13-7$

$2x=6$

Divide both sides by $2$, we get

$\frac{2x}{2}=\frac{6}{2}\therefore x=3$

(iii) $-y+36=32$

Add $-36$ both sides, we get

$-y+36-36=32-36$

$\Rightarrow -y=-4$

Multiply $(-1)$ to both sides we get

$\Rightarrow (-1)(-y)=(-1)(-4)$

$\Rightarrow y=4$

(iv) $2y-6=8$

Add $6$ to both sides we get

$2y-6+6=8+6$

$\Rightarrow 2y=14$

Divide both sides by $2$, we get

$\frac{2y}{2}=\frac{14}{2}$

$\therefore y=7$

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