SOLUTION: HI A and B each had some sweets. After A gave 26 sweets to B ,A had 38 more than B. How many more sweets did A have than B at first. Thanks

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: HI A and B each had some sweets. After A gave 26 sweets to B ,A had 38 more than B. How many more sweets did A have than B at first. Thanks      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1175523: HI
A and B each had some sweets. After A gave 26 sweets to B ,A had 38 more than B.
How many more sweets did A have than B at first. Thanks
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say that A had x sweets to begin.
Let's say that B had y sweets to begin.
We want to find and expression for the difference of A and B (A minus B) at the beginning or x-y
.
.
.
After A gives B 26 sweets, A now has x-26 sweets and B now has y%2B26 sweets.
So after the swap, the difference between A and B (A minus B) is 38.
%28x-26%29-%28y%2B26%29=38
x-26-y-26=38
x-y-52=38
x-y=38%2B52
x-y=90
So at the beginning, A had 90 more sweets than B.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x = the number of sweets that A had initially.
y = the number of sweets that B had initially.

after A gave 26 sweets to B, A had 38 more sweets than B.

the formula for that is:

x - 26 = y + 26 + 38

combine like terms to get:

x - 26 = y = 64

add 26 to both sides of the equation to get


x = y + 90

that's your answer.

A had 90 more sweets than B to start with.

to confirm this is true, let B equal any number of sweets to start with.

let B equal 500 sweets.

since A had 90 more, then A had 590.

A gives 26 sweets to B.

A now has 564 and B has 526.

A now has 564 minus 526 = 38 more sweets than B.

this confirms the solution is correct.

B could have any number of sweets to start with and the solution will be the same if A has 90 more sweets than that.