Solve by substitution: -2x+1=2y−2x+1=2y, 7x-4y-13=07x−4y−13=0
First, we can solve for yy in the first equation* by dividing both sides by 22:
-2x+1=2y−2x+1=2y
y=(-2x+1)/2y=−2x+12
We can now substitute this yy term into the second equation and solve for xx:
7x-4y-13=07x−4y−13=0
7x-4((-2x+1)/2)=137x−4(−2x+12)=13
Notice that we can cancel a 22 from the numerator and denominator:
7x-2(-2x+1)=137x−2(−2x+1)=13
7x+4x-2=137x+4x−2=13
Combining xx-terms and moving all constants to one side, we find the value of xx:
11x=1511x=15
x=15/11x=1511
To solve for yy, we plug in the xx value we just found into the first equation which we solved for yy and solve for the yy value which corresponds with x=15/11x=1511:
y=(-2x+1)/2y=−2x+12
=(-2(15/11)+1)/2=−2(1511)+12
=((-30)/11+1)/2=−3011+12
=((-30+11)/11)/2=−30+11112
=-19/22=−1922
Therefore, the solution to the linear system of equations is (15/11,-19/22)(1511,−1922)
*This was chosen because we can see that if we solve for yy in the first equation and plug into the second equation, the 1/212 resulting from dividing the first equation by 22 would cancel out with the 44 in the second equation when we substitute in (since 22 is a factor of 44). Solving for any other variable and substituting into the other equation would result in some unwanted fraction which would thus complicate the problem a bit more than necessary.
