
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

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.... :)

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! :)