A system of linear equations can be solved
algebraically using substitution or elimination.
Substitution is a method for solving a system of equations where a variable is replaced by the expression it is equal to.
Substitution is a good method for solving a system of linear equations when one of the equations can easily be solved for one of the variables.
First, use opposite operations to solve the first equation for
x.
| x + 4y - 4y | = | 37 - 4y |
| x | = | 37 - 4y |
Next, substitute
x = 37 - 4
y into the second equation, and solve for
y.
| 2(37 - 4y) - 2y | = | -6 |
| 74 - 8y - 2y | = | -6 |
| 74 - 10y | = | -6 |
| 74 - 10y - 74 | = | -6 - 74 |
| -10y | = | -80 |
| -10y ÷ (-10) | = | -80 ÷ (-10) |
| y | = | 8 |
Then, substitute
y = 8 into the first equation, and solve for
x.
| x + 4(8) | = | 37 |
| x + 32 | = | 37 |
| x + 32 - 32 | = | 37 - 32 |
| x | = | 5 |
So, x = 5 and y = 8.
Elimination is a method for solving a system of equations where the equations, or multiples of the equations, are added or subtracted in order to eliminate one variable.
Elimination is a good method for solving a system of linear equations when one of the variables has the same coefficient in both equations.
First, subtract the second equation from the first to cancel out the
x terms.
Next, substitute
y = -2 into the second equation, and solve for
x.
| 3x - (-2) | = | 14 |
| 3x + 2 | = | 14 |
| 3x + 2 - 2 | = | 14 - 2 |
| 3x | = | 12 |
| 3x ÷ 3 | = | 12 ÷ 3 |
| x | = | 4 |
So, x = 4 and y = -2.
8th Grade Math - Solve Systems of Linear Equations Lesson