One of the variables, either xx or yy , must be isolated . In this case, I will rearrange the second equation, x+3y=-28x+3y=−28 and isolate the xx variable.
x+3y=-28x+3y=−28 Now subtract 3y3y from each side of the equation
x+3y-3y=-28-3yx+3y−3y=−28−3y
x=-3y-28x=−3y−28
Now that I have isolated the xx variable in your second equation, I will substitute it in your first equation.
3x-y=-43x−y=−4
3(-3y-28)-y=-43(−3y−28)−y=−4
-9y-84-y=-4−9y−84−y=−4 collect like terms
-10y-84 = -4−10y−84=−4 Add 84 to each side
-10y-84+84=-4+84−10y−84+84=−4+84
-10y=80−10y=80 Divide by -10−10
y=-8y=−8
Now simply substitute -8−8 in for yy in either equation.
x+3y=-28x+3y=−28
x+3(-8)=-28x+3(−8)=−28
x-24= -28x−24=−28 Now add 2424 to each side of the equation
x-24+24=-28+24x−24+24=−28+24
x=-4x=−4
(-4,-8)(−4,−8)
Now there is one small problem, and that is, how do I know this is the correct solution? Since I used your second equation, x+3y=-28x+3y=−28 to calculate the value of yy, I can't use this same equation to check my answer. It would always appear to be correct. I must go to the other equation, in your case the first one, to verify that I have the correct answer. So:
3x-y=-43x−y=−4
3(-4)-(-8)=-43(−4)−(−8)=−4
-12+8=-4−12+8=−4
-4=-4−4=−4
This confirms that I have the solution correct.
