How do you solve $4 (y - 3) = 48$ $4(y-3)=48$ ?

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How do you solve $4 (y - 3) = 48$ $4(y-3)=48$ ?

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Answer 1 · 2017-06-15T17:47:53

$y = 15$ $y=15$

Explanation:

First, divide both sides of the equation by $4$ $4$ . You could also start by distributing the $4$ $4$ into the parentheses but that will only make this problem take more time.

$y - 3 = \frac{48}{4} = 12$ $y-3=48/4=12$

Add $3$ $3$ to both sides.

$y = 15$ $y=15$

Answer 2 · 2017-06-15T17:49:03

$y = 15$ $y=15$

Explanation:

$divide both sides of the equation by 4$ $"divide both sides of the equation by 4"$

$\frac{4}{4} (y - 3) = \frac{48}{4}$ $4/4(y-3)=48/4$

$\Rightarrow y - 3 = 12$ $rArry-3=12$

$add 3 to both sides$ $"add 3 to both sides"$

$ycancel(-3)cancel(+3)=12+3$

$rArry=15$

$color(blue)"As a check"$

$4(15-3)=4xx12=48=" right side"$

$rArry=15" is the solution"$

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How do you solve $4 (y - 3) = 48$ $4(y-3)=48$ ?

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