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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