How do you solve #5x + 2y = 20# and #x + 4y = 13#?

2 Answers
Apr 23, 2018

Answer:

#(x,y)to(3,5/2)#

Explanation:

#5x+2y=20to(1)#

#x+4y=13to(2)#

#"from equation "(2)" we can express x in terms of y"#

#rArrx=13-4yto(3)#

#"substitute "x=13-4y" in equation "(1)#

#5(13-4y)+2y=20larrcolor(blue)"distribute"#

#rArr65-20y+2y=20#

#rArr-18y+65=20larrcolor(blue)"subtract 65 from both sides"#

#rArr-18y=-45#

#"divide both sides by "-18#

#(cancel(-18) y)/cancel(-18)=(-45)/(-18)#

#rArry=45/18=5/2#

#"substitute "y=5/2" in equation "(3)#

#rArrx=13-(4xx5/2)=13-10=3#

#"the solution is "(x,y)to(3,5/2)#

Apr 23, 2018

Answer:

x=3, y=5/2

Explanation:

(1) #5x+2y=20#
(2) #x+4y=13#

First, multiply equation 1 by 2, this gives both equations the same y coefficient of 4. The equations are now:
(1) #10x+4y=40#
(2) #x+4y=13#

Next subtract equation 2 from 1 (equation 1 - equation 2)
This makes
#(10x-x) + (4y-4y) = (40-13)#

Simplified: #9x-0=27#

Then divide both sides by 9 to solve for x

#9/9x=27/9#
#x=3#

Then input x=3 back into equation 2

#(3)+4y=13#

Subtract 3 from both sides:
#4y=10#

Divide both sides by 4 to solve for y

#y=10/4# (simplified to #5/2#)

Therefore #x=3# and #y=5/2#