Algebra II Recipe: Solving Linear Systems Algebraically
A. Substitution Method
1. Solve one of the equations for the variable that has a 1 or -1 as a coefficient.
2. Substitute the algebraic expression into the other equation for the variable it represents.
3. Solve the equation for the like variable.
4. Substitute the value found in step 3 into the algebraic expression to find the value of the other variable.
5. 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.
Examples:
 3x + 5y = 2x = -4y - 4
 x - 2y = 32x - 4y = 7
1. Multiply one or both of the equations by some number so that one of the variables will have opposites as coefficients.
2. Add the equations to eliminate the variable having opposites as coefficients.
3. Solve the remaining equation for its variable.
4. Substitute the value found in step 3 into either one of the original equations to find the value of the other variable.
5. 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.
Examples:
 2x - 4y = 134x - 5y = 8
 7x - 12y = -22-5x + 8y = 14

G Redden

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