Question #6e4c9

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Question #6e4c9

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Answer 1 · 2016-11-13T19:43:36

Mrs. Hill's son is $15 \frac{1}{3}$ $15 1/3$ years old.

Explanation:

Let Mrs. Hill's age = $h$ $h$
Let her son's age = $s$ $s$

From the problem, the two equations are:
$h = 4 + 2 s$ $h=4+2s$
$h + s = 50$ $h+s=50$

We can solve for this using substitution because one of the variables is already isolated.
$(4 + 2 s) + s = 50$ $(4+2s)+s=50$

Simplify and solve for the son's age ( $s$ $s$ ):
$2 s + s + 4 = 50$ $2s+s+4=50$
$3 s + 4 = 50$ $3s+4=50$
$3 s = 46$ $3s=46$
$s = 15.33$ $s=15.33$ years old

Plug the son's age into one of the original equations:
$h = 4 + 2 (15.33)$ $h=4+2(15.33)$
$h = 34.67$ $h=34.67$ years old

Check:
"Mrs. Hill is 4 year older than twice her son's age. If the sum of their ages is 50, how old is her son?"
$34 \frac{1}{3}$ $34 1/3$ is 4 more than 2 times $15 \frac{1}{3}$ $15 1/3$ . $34 \frac{2}{3} + 15 \frac{1}{3}$ $34 2/3+15 1/3$ is 50.

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Question #6e4c9

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