Question 1082959
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Scott is 4 times older than Bethany. In six years from now Scott will only be 3 times older than Bethany. 
solve equations to find out how old each of them is
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<pre>
Let S be the Scott's present age and B be the Bethany's present age.

Then you have this system of two equations

S = 4B              (1)
S + 6 = 3*(B+6)     (2)

From (1), substitute the expression S = 4B into equation (2). You will get

4B + 6 = 3*(B+6),

4B + 6 = 3B + 18  ---->  B = 12.

Thus Bethany is 12 years old now.

Then Scott is 12*4 = 48 years old now.


You can easily check that this solution satisfies the condition.
</pre>

Solved.


The solution by "josgarithmetic" is WRONG.



There is a bunch of lessons on age word problems 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/age/Age-problems-and-their-solutions.lesson>Age problems and their solutions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/age/Fresh-formulation-of-a-traditional-age-problem.lesson>A fresh formulation of a traditional age problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/age/Really-intricate-age-word-problem.lesson>Really intricate age word problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/age/Selected-age-word-problems-from-the-archive.lesson>Selected age word problems from the archive</A>

in this site.


Read them and become an expert in solving age problems.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Age word problems</U>".