How do you solve #35x - 5y = 20# and #y = 7x + 4# using substitution?

2 Answers
Aug 30, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Because the second equation is already solved for #y# we can substitute #(7x + 4)# for #y# in the first equation and solve for #x#:

#35x - 5y = 20# becomes:

#35x - 5(7x + 4) = 20#

#35x - (5 * 7x) - (5 * 4) = 20#

#35x - 35x - 20 = 20#

#35x - 35x - 20 = 20#

#0 - 20 = 20#

#-20 != 20#

This indicates there are no solutions to this system of equations. Or, no solution set is the empty or null set: #{O/}#

This also indicates, if there is no solutions, that these equations represent parallel lines.

Aug 30, 2017

Answer:

"Substitute" the expression for y in the second equation into the first one.

Explanation:

#35x − 5(7x + 4) = 20#
#35x − 35x - 20 = 20#
#0 = 40#
0 does not = 40, so there is no solution to this set.

Converting both to slope-intercept form gives us:
#35x − 5y = 20# ; #-5y = -35x + 20# Dividing by -5 we get:
#y = 7x - 4#; compared to the second equation
#y = 7x + 4#

They are parallel lines (same slope) with different intercepts (4 and -4).