# help solve

• Aug 3rd 2005, 12:12 PM
Hayabusa
help solve
Use elimination to solve: x/3 + 4y = 30 and 2x - y = 5.

I need help solving this problem. If it's not to much trouble can some one help my or give me an example to help me out. I greatly apprecite the help.
• Aug 4th 2005, 01:05 AM
Cold
Hi
Use elimination to solve is a clue. It's just a fancy way of saying get rid of a variable (x or y) from one of the equations.

x/3 + 4y = 30 and 2x - y= 5.

Because the coefficients of y are nice I chose to get rid of it from one of the equations.
This step is called rearranging the equation.
2x -y =5
2x=5+y
2x-5=y

So what next? Here is the elimination part. We replace the y in x/3 + 4y=30 with (2x-5) because they are equal. (y=2x-5)

We have now have x/3+4(2x-5)=30. This looks a lot messier but we can simplify things. There are lots of ways of proceeding from here.
x/3 looks nasty - to get rid of it multiply every term (peice of the equation) by 3

x+(3*4(2x-5)=90
x+12(2x-5)=90

Next thing not to like about the equation is the brackets - essentially we want to count all the x's together. Getting rid of brackets is called multiplying out brackets. It looks strange but is just a property of the number system

12(2x-5) is identical to 12*2x - 12*5 or 24x-60 in other words we can replace 12(2x-5) by 24x - 60

we now have
x+24x -60=90
counting the x's together
25x-60=90

solving for x
25x=150
x=6

Because we know x we can find y by substuting x=6 in 2x-y=5 (for example - you could use x/3 - 4y=30 but 2x-y=5 looks easier)

12-y=5
y=7

Hope this helped - there's quite a lot of distractions in the question such as multiplying out etc.
• Aug 4th 2005, 10:56 AM
rgep
... and then we remember to check our working by showing that the solution x=6, y=7 does indeed satisfy both equations (which it does).
• Aug 4th 2005, 01:15 PM
Hayabusa
Thank you.

So I can do the same thing to this problem:

6x - y/4 = -9 and 3x + 2y = -30.
• Aug 4th 2005, 11:32 PM
Cold
Yeah. Try y=3x and 6x-2y+1=0. What does this tell you about the equations?
• Aug 5th 2005, 08:58 AM
Hayabusa
Thanks for the help.