#color(white)(...)##2x−3y=−1#
#−4x+5y=−3#
If you multiply the first equation by #-2#, the coefficients of the #x# terms in both equations will be the same.
Then you can eliminate the #x# terms by subtracting one equation from the other so that the #x# terms go to zero.
Then it's easy to solve an equation in a single unknown.
#color(white)(....................)#. . . . . . . . . . . . .
1) Multiply all the terms on both sides of the first equation by #-2# to make the coefficients of the #x# terms the same.
After you multiply, you will get this:
#-4x+6y=2#
2) Subtract one of the equations from the other to make the #x# terms drop out
#color(white)(....)#(#-4x+6y= 2#)
#-#(#−4x+5y=##"-"# #3#)
3) Clear the parentheses by distributing the minus sign
#-4x+6y=2#
#+4x-5y=3#
4) Add the equations and let the #x# terms go to 0
#y = 5# #larr# answer for #y#
#color(white)(....................)#. . . . . . . . . . . . .
Use the value (#y = 5#) to find the value of #x#
1) Sub in #5# in the place of #y# in one of the equations
#2x−3 (y)=−1#
#2x−3 (5)=−1#
2) Clear the parentheses
#2x - 15 = - 1#
3) Add #15# to both sides to isolate the #2x# term
#2x = 14#
4) Divide both sides by #2# to isolate #x#
#x = 7# #larr# answer for #x#
Answer:
#x = 7#
#y = 5#
