Solve by substitution: #-2x+1=2y#, #7x-4y-13=0#
First, we can solve for #y# in the first equation* by dividing both sides by #2#:
#-2x+1=2y#
#y=(-2x+1)/2#
We can now substitute this #y# term into the second equation and solve for #x#:
#7x-4y-13=0#
#7x-4((-2x+1)/2)=13#
Notice that we can cancel a #2# from the numerator and denominator:
#7x-2(-2x+1)=13#
#7x+4x-2=13#
Combining #x#-terms and moving all constants to one side, we find the value of #x#:
#11x=15#
#x=15/11#
To solve for #y#, we plug in the #x# value we just found into the first equation which we solved for #y# and solve for the #y# value which corresponds with #x=15/11#:
#y=(-2x+1)/2#
#=(-2(15/11)+1)/2#
#=((-30)/11+1)/2#
#=((-30+11)/11)/2#
#=-19/22#
Therefore, the solution to the linear system of equations is #(15/11,-19/22)#
*This was chosen because we can see that if we solve for #y# in the first equation and plug into the second equation, the #1/2# resulting from dividing the first equation by #2# would cancel out with the #4# in the second equation when we substitute in (since #2# is a factor of #4#). 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.
