How do you solve the system x/3+y/5=2x3+y5=2 and x/3 - y/2= -1/3x3y2=13?

1 Answer
Mar 25, 2016

I did this in a hurry so you need to check the calculations!
y=40/21y=4021

x=51/4x=514

Explanation:

My first reaction was to consider getting rid of the fractions. That was, until I spotted the coefficients of xx are the same in both equation.

Given:

x/3+y/5=2x3+y5=2..............................(1)

x/3-y/2=-1/3x3y2=13...........................(2)

Subtract equation (2) from (1)

0+y/5+y/2=2+1/30+y5+y2=2+13

(2y+5y)/10=4/3 2y+5y10=43

7y=40/37y=403

y=40/21y=4021

Now all you have to do is substitute

Substitute in equation (1) giving

x/3+(1/5xx40/21)=2x3+(15×4021)=2.........................(1_a)

x/3 = 2-8/21 = 34/8x3=2821=348

x=(3xx34)/8 =12 3/4 =51/4x=3×348=1234=514