How do you solve #y=7x-1# and #y=-x+14# using substitution?

2 Answers
Apr 26, 2018

Answer:

The solution is #(15/8,97/8)# or #(1.875,12.125)#.

Explanation:

Solve the system:

#"Equation 1:"# #y=7x-1#

#"Equation 2:"# #y=-x+14#

Both linear equations are in slope-intercept form. The solution to the system is the point that both lines have in common, the point of intersection. Substitution will be used to solve the system.

Both equations are set equal to #y#. Substitute #7x-1# from Equation 1 for #y# in Equation 2 and solve for #x#.

#y=-x+14#

#7x-1=-x+14#

Add #x# to both sides.

#x+7x-1=14#

Simplify.

#8x-1=14#

Add #1# to both sides.

#8x=15#

Divide both sides by #8#.

#x=15/8# or #1.875#

Substitute #15/8# for #x# in Equation 1 and solve for #y#.

#y=7x-1#

#y=7(15/8)-1#

#y=105/8-1#

Multiply #1# by #8/8# to get a denominator of #8#.

#y=105/8-1xx8/8#

#y=105-8/8#

Combine the numerators.

#y=(105-8)/8#

#y=97/8# or #12.125#

The solution is #(15/8,97/8)# or #(1.875,12.125)#.

graph{(y-7x+1)(y+x-14)=0 [-9.455, 10.545, 5.48, 15.48]}

Apr 26, 2018

Answer:

#(x, y): (15/8, 97/8)#

Explanation:

#eq_1 : y=7x-1#
#eq_2 : y=-x+14#

Use #7x - 1# for y in #eq_2#:
#7x - 1 = -x + 14#
#8x = 15#
#x = 15/8#

#y=7x-1#
#y = 7xx15/8-8/8#
#y = 97/8#