1. ## 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

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.

2. Originally Posted by alwaysalillost
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

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

---------------------------------
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

----------------------------------

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

------------------------------------

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

------------------------------------------------

Hope this helps....

3. 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.

4. ~Thanks so much for the explaination Janvdl!~

5. You're very welcome.