Question 333745
My algebra teacher gave us this problem, I don't know how to start solving it. Please help me.
<pre><b>
Let X = Marty's age now

Let Y = Mother's age now

Then 

Marty's age five years age = (X-5)

Mother's age five years ago = (Y-5)

1/3 of Mother's age now = {{{1/3}}}{{{Y}}}

1/5 of Mother's age five years ago = {{{1/5}}}{{{(Y-5)}}}
</pre></b>
>>...Marty is 1/3 as old as her mother. 
<pre><b>
Marty's age = 1/3 of Mother's age now
  
     X      =  {{{1/3}}}  {{{""*""}}}   {{{Y}}}       
</pre></b>
>>...Five years ago she (Marty) was only 1/5 of the age of her mother...<< 
<pre><b>
Marty's age five years ago = 1/5 of Mother's age five years ago 

         (X-5)             =  {{{1/5}}}  {{{""*""}}} {{{(Y-5)}}}

     
So we have these two equations 

{{{system(X=(1/3)Y,(X-5)=(1/5)*(Y-5))}}}

Multiply both sides of the first equation by 3 to clear of fractions.
Multiply both sides of the first equation by 5 to clear of fractions.

{{{system(red(3)X=red(3)(1/3)Y,red(5)(X-5)=red(5)(1/5)*(Y-5))}}}

Canceling clears fractions:

{{{system(red(cross(3))X=red(cross(3))(1/cross(3))Y,red(5)(X-5)=red(cross(5))(1/cross(5))*(Y-5))}}}

{{{system(3X=Y,red(5)(X-5)=(Y-5))}}}

Simplifying further:

{{{system(3X=Y,5X-25=Y-5)}}}

Simplifying further:

{{{system(3X=Y,5X=Y+20)}}}

Since the first equations tells us that Y = 3X we can replace the Y
in the second equation 

{{{5X=Y+20)}}}

by 3X

{{{5X=3X+20}}}

{{{2X=20}}}

{{{X=10}}}

So Marty is 10 years old.  However we need to find her
Mothers age so we can check.  We use

Y = 3X

Y = 3(10)

Y = 30

So her mother is 30.

Let's check:

Marty is 1/3 as old as her mother.

Indeed 10 is 1/3 of 30

Five years ago she was only 1/5 of the age of her mother

Five years ago se was 5 and he mother was 25, and indeed
5 is 1/5 of 25.

So we know we are right.

Edwin</pre>