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Question 221033: Four years ago Katie was twice as old as Anne was then. in 6 years, Anne will be the same age that katie is now. How old is each now ?
I have two equattions of Katie= X, y= Anne
2X-4= y
x+6 = 2y - 4
Answer by drj(1380)   (Show Source):
You can put this solution on YOUR website! Four years ago Katie was twice as old as Anne was then. In 6 years, Anne will be the same age that Katie is now. How old is each now ?
I have two equations of Katie= X, y= Anne
2X-4= y
x+6 = 2y - 4
You're in the right track and good try in getting two linear equations....I'll check your work by the following steps.
Step 1. Let x be Katie's age and y be Anne's age.
Step 2. Let x-4 be Katie's age 4 years ago and y-4 be Anne's age four years ago.
Step 3. Then, x-4= 2(y-4) since four years ago Katie was twice as old as Anne was then. Simplifying yields x-2y=-4
Step 4. Let y+6 be the age of Anne.
Step 5. Then y+6=x since in 6 years, Anne will be the same age that Katie is now. Simplifying yields x-y=6
Step 6. The system of equations from Steps 3 and 5 are
x-2y=-4 Equation A
x-y-=6 Equation B
OR multiplying Equation B by -1 yields
x-2y=-4 Equation A
-x+y=-6 Equation B1
Adding yields
x-x-2y+y=-4-6
or y=10 and x=10+6=16
Substituting x=16 and y=10, into Steps 3 and 5 yields true statements.
Step 7. ANSWER: Kate is 16 years old and Anne is 10 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
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