S.O.S. Mathematics CyberBoard

Dear Friends,

How should I set up this word problem?

2. Twice the sum of two numbers exceeds 3 times their difference by 8. While 1/2 the sum is 1 more than the difference. What are the numbers?

I did this: 2(x + y)/2 – 3(x-y) = 8

Then I did this:
(x+x1)/2=(x-x1)-1
(x+ x1)/2=(x- x1)-1
½x+½x1 = (x- x1)-1
½x-x+½x1+ x1= -1
-½x+3/2x1= -1
x = -1

I think I have to go and learn maths again, lol.

What am I doing wrong?

Mike

Twice the sum of two numbers exceeds 3 times their difference by 8. While 1/2 the sum is 1 more than the difference. What are the numbers?

How about this: and

These can be rewritten into the much more user friendly "standard form" .
Please show us what you can do with that.

_________________
"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)

Just to clarify the transformation a bit:

• Twice the sum of two numbers exceeds 3 times their difference by 8.
• Two times (the sum of two numbers) = three times (their difference), plus 8.
• 2(x+y) = 3(x-y)+8

Now, the difference of x and y is normally |x-y|, but in this case we can just define x to be the bigger one and replace that with plain x-y.

Mike, 2(x + y) = 2x + 2y

RULE: a(x + y) = ax + ay ;
in other words, multiply what's inside brackets by what's outside...got that?

_________________
I'm not prejudiced...I hate everybody equally!