Algebra II Recipe: Solving Linear Systems Algebraically
By G Redden
- Solve one of the equations for the variable that has a 1 or -1 as a coefficient.
- The variable’s value will be an algebraic expression – BOX IT.
- Substitute the algebraic expression into the other equation for the variable it represents.
- Solve the equation for the like variable.
- Substitute the value found in step 3 into the algebraic expression to find the value of the other variable.
- When solving the equation in step 3 – if the variable terms cancel out on both sides:
- The answer is IMS, if a true statement remains.
- The answer is NO SOLUTION, if a false statement remains.
1.
3x + 5y = 2
x = -4y - 4
x = -4y - 4
2.
x - 2y = 3
2x - 4y = 7
2x - 4y = 7
- Multiply one or both of the equations by some number so that one of the variables will have opposites as coefficients.
- Add the equations to eliminate the variable having opposites as coefficients.
- Solve the remaining equation for its variable.
- Substitute the value found in step 3 into either one of the original equations to find the value of the other variable.
- When adding the equations in step 2 – if both variables cancel:
- The answer is IMS, if a true statement remains.
- The answer is NO SOLUTION, if a false statement remains.
3.
2x - 4y = 13
4x - 5y = 8
4x - 5y = 8
4.
7x - 12y = -22
-5x + 8y = 14
-5x + 8y = 14