How do you solve #0.5x+0.25y=36# and #y+18=16x# using substitution?

1 Answer
Mar 5, 2016

Answer:

#x = 5#
#y = 62#

Explanation:

In either equation, express one of the variables in terms of the other terms.

#[1] 0.5x + 0.25y = 36#
#[2] y + 18 = 16x#

If we solve for #y# in #[2]#, we have

#[3] => y = 16x - 18#

Now, substitute this relation into #[1]#

#[1] 0.5x + 0.25y = 36#

#=> 0.5x + 0.25(16x - 18) = 18#

Multiplying the entire equation by 4, we have

#=> 4(0.5x + 0.25(16x - 18) = 18)#

#=> 2x + 1(16x - 18) = 72#
#=> 2x + 16x - 18 = 72#

#=> 18x = 90#
#=> x = 5#

Using #[3]# (or #[1]# or #[2]#), solve for #y#

#y = 16x - 18#
#=> y = 16(5) - 18#

#=> y = 80 - 18#

#=> y = 62#