x + y = 5 and y = 3x - 1
x - y = -3 2x + y = 14
Solve using substitution.
First, isolate one variable.
x + y = 5
x = 5 - y
Then, plug its expression into the other equation and solve.
x - y = -3
5 - y - y = -3
5 - 2y = -3
-2y = -8
y = 4
Next, plug your new value into one of the original equations to solve for the second variable.
x + y = 5
x + 4 = 5
x = 1
Finally, check work with substitution.
x + y = 5 => 1 + 4 = 5 <--True
x - y = -3 => 1 - 4 = -3 <--True
Next.
y = 3x - 1
2x + y = 14
2x + y = 14
2x + (3x - 1) = 14
2x + 3x - 1 = 14
5x - 1 = 14
5x = 15
x = 3
y = 3x - 1
y = 3(3) - 1
y = 9 -1
y = 8
y = 3x - 1 => 8 = 3(3) - 1 => 8 = 9 - 1 <--True
2x + y = 14 => 2(3) + 8 = 14 => 6 + 8 = 14 <--True
Answer:
4.
x = 1
y = 4
5.
x = 3
y = 8
