Thread: Substitution Method

Solve and check each of the following problems using the substitution method.

1. 3x + y = 5
-2x + 3y = 4

2. y = 5x
2x + y = 28

3. x + 3y = 23
3x - 2y = 3

4. x = 1/5y
3x + 2y = 26

I need help in figuring out the answers to these problems. If you could please explain to me how to solve it I'd appreciate it very much.

1. y = 5 - 3x
Set this into (-2x + 3y = 4)

-2x + 3(5 - 3x) = 4
-2x + 15 - 9x = 4
-11x = -11
x = 1

Set x = 1 into (y = 5 - 3x)
y = 5 - 3(1)
y = 2

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2. y = 5x
Set y = 5x into the 2nd equation (2x + y = 28)
2x + 5x = 28
x = 4
Set x = 4 into the 1st equation
y = 5(4) = 20

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3. x + 3y = 23

x = 23 - 3y
Set into 2nd equation(3x - 2y = 3)
3(23 - 3y) - 2y = 3
69 - 9y - 2y = 3
-11y = -66
Thus y = 6
Set y = 6 into 1st equation
x + 3(6) = 23
x = 5

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4. x = 1/5y
3x + 2y = 26

Set (x = 1/5y) into 2nd equation

3(1/5y) + 2y = 26
3/5y + 2y = 26
3y + 10y + 130
13y = 130
y = 10

Set y = 10 into 1st equation

x = 1/5 x 10
x = 2

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Hope this helps....

Never insert a calculated value into the same equation you got its value from. You'll only get an answer stating that the LHS = RHS. Which is true, but its not what you want.

Thanks so much Janvdl!