SOLUTION: I am having problems with the following system of equations. The elimination method is supposed to be used, and the solution is to be in the form of integers or fractions. .03x

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I am having problems with the following system of equations. The elimination method is supposed to be used, and the solution is to be in the form of integers or fractions. .03x      Log On


   

Question 133058: I am having problems with the following system of equations. The elimination method is supposed to be used, and the solution is to be in the form of integers or fractions.
.03x-0.2y=4
.04x+0.5y=63/19
Thanks!
Stephanie
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you are being asked to eliminate rather than use substitution.
Given:
.03x-0.2y=4 Equation A
.04x%2B0.5y=63%2F19 Equation B
In order to use elimination, you want to get the coefficients of one of the variables (let's say y in this case) to be the same. Then you can simply add/substract one equation from the other to eliminate that variable. Once you have a single variable left, you can solve for it easily enough.
So let's manipulate equation A above. Let's multiply it by 5.
+0.15x+-+y+=+20
Now manipulate equation B. Let's multiply it by 2.
+0.8x+%2B+y+=+126%2F19
Then add these two equesitons to get:
+0.23x+=+20%2B+126%2F19+
solve for x, then use that to solve for y in any of the equations above.

So now the question?? How did we know to use 5 and 2? Great question. We need to find a way to make the coefficients of one of the variables end up being equal in order to add/subtract them. You can look at Equation A and see that if you multiply by 5 it becomes 1. Then pick a number to multiply equation B to also se the y coefficient to 1 (in this case that number is 2).