How do you solve #x = 4y# and #2x + 3y = 44# using substitution?

2 Answers
Jun 1, 2018

You have #x = 4y#, so, what you do is, substitute #4y# wherever you see #x# in the second equation. This sequence looks like:

#2x + 3y = 44#

#2(4y) + 3y = 44#

#8y + 3y = 44#

#11y = 44#

#y = 44/11 = 4#

...now that you know that y = 4, you can plug this in to either equation and solve for x.

#x = 4y = 4(4)#

#x = 16#

CHECK YOUR WORK. Feed these values into each equation to verify that they are actually solutions.

#x = 4y#
#16 = 4 * 4# CHECK

#2x + 3y = 44#

#(2 * 16 ) + (3 * 4) = 32 + 12 = 44# CHECK

GOOD LUCK

Jun 1, 2018

Answer:

#(x,y)to(16,4)#

Explanation:

#x=4yto(1)#

#2x+3y=44to(2)#

#color(blue)"substitute "x=4y" into equation "(2)#

#2(4y)+3y=44#

#8y+3y=44#

#11y=44rArry=44/11=4#

#"substitute "y=4" into equation "(1)#

#x=4xx4=16#

#"point of intersection "=(16,4)#