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# Solving One-Step Equations

## Solving One-Step Equations 1.3 - Solution

a
To solve the equation, we need to isolate $x$ on one side of the equation. Subtracting $2$ from both sides will "cancel" it from the right-hand side.
$6=x+2$
$6-2=x+2-2$
$4=x$
$x=4$
b
To solve the equation we have to isolate $x$ on one side. By adding $11$ to both sides, we can cancel out $\text{-} 11$ and isolate $q.$
$q-11=\text{-}9$
$q-11+11=\text{-}9+11$
$q=2$
c
To solve the equation we have to isolate $y$ on one side of the equation. By dividing both sides of the equation by $\text{-}7,$ we can remove the coefficient on the left-hand side.
$\text{-}7y=28$
$\dfrac{\text{-}7y}{\text{-}7}=\dfrac{28}{\text{-}7}$
$y=\dfrac{28}{\text{-}7}$
$y=\text{-}\dfrac{28}{7}$
$y=\text{-}4$
d
By multiplying both sides of the equation by $2,$ we can eliminate the denominator and isolate $z.$
$14=\dfrac{z}{2}$
$14\cdot 2=\dfrac{z}{2}\cdot 2$
$14\cdot 2=z$
$28=z$
$z=28$