How do you solve 2x+6y=32 and 5x-2y=12 using substitution?

2 Answers
Mar 11, 2018

therefore color(red)(x=4 and y=4

Explanation:

2x+6y=32

=> x + 3y = 16

Multiplying both sides by 2

x=16-3y

Given that,

5x−2y=12

Replacing x=16-3y

5(16-3y)-2y=12

80-15y-2y=12

80-17y=12

17y=80-12

17y=68

y=68/17

color(magenta)(y=4

therefore x=16-3y

x=16-3xx4

x=16-12

color(magenta)(x=4

therefore color(red)(x=4 and y=4

~Hope this helps! :)

Mar 11, 2018

x=4, y=4

Explanation:

2x+6y=32
5x-2y=12

In order to use substitution, rearrange one of the equations

2x+6y=32
2x= -6y+32 (Subtracted 6y from both sides)
x=-3y+16 (Divided both sides by 2)

Now that we rearrange the equation, we plug it into 5x-2y=12

5(-3y+16)-2y=12
-15y+80-2y=12 (Distributed 5 to the terms in bracket)
-17y+80=12 (Combined like terms)
-17y = -68 (Subtract 80 from both sides)
y=4 (Divided both sides by -17)

Now that we have the y term, we can plug it into either equation to find x

2x+6(4) = 32
2x+24=32 (Multiplied 6*4)
2x= 8 (Subtracted 24 from both sides)
x=4 (Divided both sides by 2)

Check your answer y plugging in y=4 and x=4 into either equation.