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

#### Solution

i) $7x=28$
Divide both sides by $7$
$\dfrac{7x}{7}=\dfrac{28}{7}$
$\Rightarrow x=4$

(ii) $2x+7=13$
Add $-7$ to both sides,
$2x+7-7=13-7$
$2x=6$
Divide both sides by $2$, we get
$\dfrac{2x}{2}=\dfrac{6}{2} \therefore x=3$

(iii) $-y+36=32$
Add $-36$ both sides, we get
$-y+36-36=32-36$
$\Rightarrow -y=-4$
Multiply $\left(-1\right)$ to both sides we get
$\Rightarrow \left(-1\right)\left(-y\right)=\left(-1\right)\left(-4\right)$
$\Rightarrow y=4$

(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
$\dfrac{2y}{2}=\dfrac{14}{2}$
$\therefore y=7$

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