# Need algebra help.

• Jul 24th 2005, 11:12 AM
Cyberman
Need algebra help.
{ 3/4x + 2/3y = 7
{ 3/5x - 1/2y = 18

I'm suppose to find out what x and y are.

the answer from the book is:

(20,-12)

I think my problem here is trying to take the fractions and turn them into regular numbers. Maybe that's not what I need to do. I did this problem a few times and I got these whacked out, far fetched answers for x and y so I gave up.

I probably am missing something and solving this is simpler than I think.
I'm guessing the substitution "injection" method would be used here.
• Jul 24th 2005, 07:44 PM
Jamilworm
It is a good idea to get rid of those fractions, because they can get confusing when you divide and miltiply by them. The way to get rid of them is to multiply the whole equation by the least common multiple of all of the denominators (I think I said that right). Anyway, just find the smallest number that each denominator can divide evenly into. This will clean up the equations.

3/4x + 2/3y = 7 : This one you can multiply everything by 12, giving:
9x + 8y = 84

3/5x - 1/2y = 18: multiply by 10:
6x - 5y = 180

You might be able to do it now, but just in case...
These people are always talkign about subtracting equations from each other, but with only 2 variables, i find it easiest to take one equation and solve for one of the variables in terms of the other one. Then plug that result into the other equation, to get a numerical value for one variable.

ie. solve for x in the second equation:
x = 30 + 5/6y.
now plug that into the first equation and simplify:
9(30 + 5/6y) + 8y = 84 --> 270 + 15/2y + 8y = 84 --> 31/2y = -186 --> y = -12

and since x = 30 + 5/6y --> x = 30 + 5/6(-12) = 30 -10 = 20.

If you don't already know the answers, don't forget to plug your answers into the initial equations to make sure they work out.