Math 1 · Unit 6 · Lesson 2 · part 3 of 4
The repeatable procedure
1.
The repeatable procedure
- 1Solve one equation for one variableIf an equation is already y = ... or x = ..., use it. If not, pick the equation and the variable that come free most easily (a lonely x or y with coefficient 1).
- 2Substitute that expression into the other equationReplace the variable with the expression. Use parentheses. Now there is one variable.
- 3Solve the one-variable equationUnit 1: distribute, combine like terms, add opposites, multiply by the reciprocal.
- 4Back-substitute to find the other variablePut the number you found into the solved equation from step 1.
- 5Check the pair in both equationsBoth true, or something slipped.
Worked: y = x − 3 and 2x + y = 9
1. The first is already solved for y. 2. 2x + (x − 3) = 9. 3. 3x − 3 = 9, so 3x = 12, so x = 4. 4. y = 4 − 3 = 1. 5. Check: 1 = 4 − 3, true; 2(4) + 1 = 9, true. (4, 1).
Worked: x + 2y = 8 and 3x − y = 3
1. The lonely x: x = 8 − 2y. 2. 3(8 − 2y) − y = 3. 3. 24 − 6y − y = 3, so 24 − 7y = 3, so −7y = −21, so y = 3. 4. x = 8 − 2(3) = 2. 5. Check: 2 + 6 = 8, true; 6 − 3 = 3, true. (2, 3).
Use parentheses3(8 − 2y) means the whole expression is multiplied by 3. Distribute (Unit 1, Lesson 2).
Back-substitute into the easy oneThe equation from step 1 is already solved, so it gives the other variable in one move.
When substitution fits bestWhen a variable is already alone, or has coefficient 1 so it comes free in one step.