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

2 Answers
Apr 23, 2018

(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

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