Question 1175640
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There are dozens of ways to set this problem up for solving.<br>
I will show you a path to the solution that is a bit unusual but that "works" well for me.<br>
You will probably get responses from other tutors that show more traditional algebraic solution methods.<br>
"In six years Janelle will be three-eighths as old as her mother."<br>
With that statement, most ways of setting up the problem would involve working with fractions.  I find it easier to avoid fractions, as the computations are a bit easier and faster without them.<br>
So my start on the problem is this:<br>
Let the ages of Mrs. Andrew and her daughter 6 years from now be 8x and 3x.<br>
Then their current ages are 8x-6 and 3x-6.<br>
Mrs. Andrew's age is 2 less than 4 times Janelle's age:<br>
{{{8x-6 = 4(3x-6)-2}}}
{{{8x-6 = 12x-24-2 = 12x-26}}}
{{{20 = 4x}}}
{{{x = 5}}}<br>
ANSWERS:
Mrs. Andrew's age is 8x-6 = 40-6 = 34
Janelle's age is 3x-6 = 9<br>