Question 215999
Tom is half as old as Ellen was three years ago.  
In 10 years, Ellen will be one-and-one-half times as old as Tom.

Would you please show me the formula to use?  And the answer?  

I probably need more info but here is the problem set-up.


Step 1.  Let {{{x}}} be Tom's current age and let {{{y}}} be Ellen's age


Step 2.  Let {{{x-3}}} be Tom's age three years ago and {{{y-3}}} be Ellen's age three years ago


Step 3.  Let {{{y-3=2(x-3)}}} be Ellen's age three years ago since Tom is half of Ellen's age three years ago.  Simply this equation as follows


{{{y-3=2(x-3)}}}


{{{y-3=2x-6}}}


Add 6-y to both sides of the equation to isolate x and y terms on one side and numbers on the right side


{{{y-3+6-y=2x-6+6-y}}}


{{{3=2x-y}}}


{{{2x-y=3}}}


Step 4.  Let {{{y+10=3(x+10)/2}}} be Ellen's age in ten years since Ellen will be one and one half times as old as Tom where {{{1.5=3/2}}}, x+10 is Tom's age in ten years, and y+10 is Ellen's age in ten years.


Multiply by 2 to both sides of the equation to get rid of denominator


{{{2(y+10)=2*3(x+10)/2}}}


{{{2y+20=3x+30}}}


Subtract 3x-20 to both sides of the equation


{{{2y+20-3x-20=3x+30-3x-20}}}


{{{-3x+2y=10}}}




Step 5.  Now we have a linear system of equations from Steps 3 and 4 and is shown below:

{{{2x-y=3}}}
{{{-3x+2y=10}}}


Step 6.  Now solve the linear system of equations in Step 5 using substitution.


*[invoke linear_substitution "x", "y", 2,-1, 3, -3, 2, 10]


So x=16 and y=29.


Check the ages three ago


{{{x-3=16-3=13}}} and {{{y-3=29-3=26}}} where Ellen is twice the age of Tom three years ago...so this is a true statement.


Check the ages ten years from now.


{{{x+10=16+10=26}}} and {{{y+10=29+10=39}}} or {{{39=3*26/2=39}}} which is another true statement.


Step 7.  So Tom is 16 years old and Ellen is 29 years old.


I hope the above steps were helpful.


For FREE Step-By-Step videos in Introduction to Algebra, please visit 

http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit 

http://www.FreedomUniversity.TV/courses/Trigonometry.


Good luck in your studies!


Respectfully,
Dr J