1. Solving Systems of Equations Algebraically Chapter 3.2

2. Alternatives to Graphing Sometimes, graphing systems is not the best way to go – Lines don’t intersect at a discernable point – Don’t have enough room to graph – Etc. Alternate methods of solving systems – Substitution – Elimination

3. Substitution Substitution: – One equation is solved for one variable in terms of the other. This expression can be substituted for the variable in the other equation.

4. Example 1 Solve either equation for one variable Substitute this expression into the other equation – Now, there should only be one variable After solving for one variable, plug it into either equation to solve for other variable x + 4y = 26 x - 5y = -10 x = 5y – 10 5y – 10 + 4y = 26 9y – 10 = 26 9y = 36 y = 4 x + 4y = 26 x + 4(4) = 26 x + 16 = 26 x = 10 Solution: (10, 4)

5. Example 2 Solve either equation for one variable Substitute this expression into the other equation – Now, there should only be one variable After solving for one variable, plug it into either equation to solve for other variable 2x + y = 4 3x + 2y = 1 y = -2x + 4 3x + 2(-2x + 4) = 1 3x - 4x + 8 = 1 -x + 8 = 1 -x = -7 x = 7 2x + y = 4 2(7) + y = 4 14 + y = 4 y = -10 Solution: (7, -10)

6. Elimination Elimination: – Eliminate one of the variables by adding or subtracting the two equations together

7. Example 3 Instead of using Substitution, you can subtract one equation from the other – Subtract the 2 nd from the 1 st After solving for one variable, plug it into either equation to find the other x + 2y = 10 x + y = 6 0 + y = 4 y = 4 x + y = 6 x + 4 = 6 x = 2 Solution: (2, 4)

8. Example 4 Instead of using Substitution, you can add one equation to the other After solving for one variable, plug it into either equation to find the other 2x + y = 5 3x - y = 20 5x + 0 = 25 x = 5 2x + y = 5 2(5) + y = 5 10 + y = 5 y = -5 Solution: (5, -5)

9. Elimination with Multiplication When one variable cannot be easily eliminated using simple addition or subtraction, multiply one or both equations by constants so that a variable CAN be eliminated.

10. Example 5 Simple addition or subtraction isn’t going to help here. Decide which variable to eliminate – Let’s do x Now subtract 2 nd from 1 st Use one variable to solve for the other 2x + 3y = 12 5x - 2y = 11 10x + 15y = 60 10x - 4y = 22 0 + 19y = 38 y = 2 x5 x2 2x + 3y = 12 2x + 3(2) = 12 2x + 6 = 12 2x = 6 x = 3 Solution: (3, 2)

11. Example 6 Simple addition or subtraction isn’t going to help here. Decide which variable to eliminate – Let’s do h Now add together Use one variable to solve for the other 2g + h = 6 3g - 2h = 16 4g + 2h = 12 3g - 2h = 16 7g + 0 = 28 g = 4 x2 2g + h = 6 2(4) + h = 6 8 + h = 6 h = -2 Solution: (4, -2)

12. Inconsistent and Dependent Systems If you add or subtract two equations in a system and the result is an equation that is never true, then the system is inconsistent and it has no solution. – Examples 1 = 2 -1 = 1 If the result is an equation that is always true, then the system is dependent and has infinitely many solutions. – Examples 1 = 1 9 = 9

13. Example 7 (substitution) Solve either equation for one variable Substitute this expression into the other equation – Now, there should only be one variable x + 3y = 8 1/3 x + y = 9 x = -3y + 8 1/3(-3y + 8) + y = 9 -y + 8/3 + y = 9 8/3 = 9 Not True No Solutions!!

14. Example 8 (substitution) Solve either equation for one variable Substitute this expression into the other equation – Now, there should only be one variable 2a - 4b = 6 -a + 2b = -3 a = 2b + 3 2(2b + 3) - 4b = 6 4b + 6 - 4b = 6 6 = 6 Always True Infinitely Many Solutions!!

15. Example 9 (elimination) Simple addition or subtraction isn’t going to help here. Decide which variable to eliminate – Let’s do x Now add together 4x - 2y = 5 -2x + y = 1 -4x + 2y = 2 4x - 2y = 5 0 + 0 = 7 Not True x2 No Solutions