How do you solve the system #2x-y= 7# and #3x-4y= -7#?

1 Answer
May 19, 2018

Answer:

#x = 7 and y = 7#

Explanation:

#2x - y = 7 - - - eqn1#

#3x - 4y = -7 - - - eqn2#

Multiply #eqn1# by #3# and #eqn2# by #2#

#3 (2x - y = 7)#

#2 (3x - 4y = -7)#

#6x - 3y = 21 - - - eqn3#

#6x - 8y = -14 - - - eqn4#

Subtracting #eqn4# from #eqn3#

#(6x - 6x) + (-3y - (-8y)) = 21 - (-14))#

#0 - 3y + 8y = 21 + 14#

#5y = 35#

#y = 35/5#

#y = 7#

Substiting the value of #y# into #eqn1#

#2x - y = 7 - - - eqn1#

#2x - 7 = 7#

#2x = 7 + 7#

#2x = 14#

#x = 14/2#

#x = 7#

Therefore;

#x = 7 and y = 7#