5. Video Break!!!!! Click on this link to watch videos on solving equations.

6. Recall General Rules Order of Operations: PEMDAS Multiplication/Division is done in order, left to right Addition/Subtraction is done in order, left to right Solving Equations First, you must know what you are solving for so you can isolate it. To do that: Take care of any exponents/FOIL or distribution/simplification Get common denominators if necessary Combine like terms on each side of the equal sign Addition/Subtraction across = is done next to isolate the variable Multiplication/Division across =is done last and the variable should now be isolated

7. Gizmos Gizmo: Modeling 2-Step Equations Gizmo: Modeling 1-Step Equations B Gizmo: Modeling 1-Step Equations A Gizmo: Solving 2-Step Equations

11. Example 2 Example 1: Solve the equation 5(h – 2) = -4(3-h). 5(h - 2) = -4(3 – h) 5h – 10 = -12 + 4h +10 +10 5h = -2 + 4h 5h = -2 + 4h -4h - 4h h = -2 Check: 5( h - 2) = -4(3 – h ) 5(( -2 ) – 2) = -4(3 – ( -2 )) 5( - 4) = -4 ( 3 + 2) -20 = - 4 (5) -20 = -20 Since both sides of the equation yield the same result, we know that our answer is correct! Solution: We have a couple of choices as to how to solve this equation. You may notice that it is not in the form introduced earlier , but it still a linear equation in one variable, since it can be written in the form described above. Our goal is to isolate the variable h , so that it appears on one side of the equation, and its value appears on the other. We will need to distribut e on both sides of the equation before we can do that.

12. Example 1 Example 1: Solve the equation Remember: Solving Equations First, you must know what you are solving for so you can isolate it. To do that: Take care of any exponents/FOIL or distribution/simplification Get common denominators if necessary Combine like terms on each side of the equal sign Addition/Subtraction across = is done next to isolate the variable Multiplication/Division across =is done last and the variable should now be isolated

13. Example 1: Solution -45 -45 -10m -10m 140 140

14. Check: We find that both sides of the equation give us the same result when we plug our answer in, which means that we obtained the correct result!

15. Practice Examples Example 2: Example 3: Solve the equation 3p + 2 = 0 . Solve the equation -7m – 1 = 0. Solutions on next slide. Solve these on your own first. Example 4: Solve the equation 14z – 28 = 0.

16. Practice Examples Answers Example 2: Example 3: Solve the equation 3p + 2 = 0 . Solve the equation -7m – 1 = 0. Example 4: Solve the equation 14z – 28 = 0. Check: Check: Check:

17. More Practice Examples Example 5: Example 6: Solve the equation . Solve the equation . Solve these on your own first. Solutions on next slide.

18. More Practice Examples - Answers Example 5: Example 6: Solve the equation . Solve the equation .

19. More Practice Examples Example 7: Example 8: Solve the equation . Solve the equation . Solve these on your own first. Solutions on next slide.

20. Example 7: Example 8: Solve the equation . Solve the equation . More Practice Examples - Answers

21. More Practice Examples Example 9: Example 10: Solve the equation for m . . Solve the equation for . Solve these on your own first. Solutions on next slide.

22. More Practice Examples Example 9: Example 10: Solve the equation for m . . Solve the equation for .

23. More Practice Examples Example 11: Solve the equation for x. Obviously, , so the answer is No Solution No Solution

25. Solution a): We want to isolate C on one side of the equation. So, we apply the following operations on our original equation. -32 -32 or Now we have an equation that allows us to compute degrees Celsius if we knew degrees Fahrenheit.

26. Solution b): We plug 98 in for F and solve for C . -32 -32 9 9 So, 98ºF is approximately 36.67ºC.

27. Example: When you buy a new car, they say that the value of the car depreciates as soon as you drive it off the lot! Accountants use the following equation to measure depreciation of assets: where … D is the depreciation of the asset per year, C is the initial cost of the asset, S is the salvage value, and L is the asset’s estimated life. a. What is the salvage value of a machine that cost a company $40,000 initially, has an annual depreciation of $3000, and an estimated life of 10 years? b. Solve the original equation for S , the salvage value, in general.

28. Solution a): We plug 30,000 in for C , 2000 for D , and 10 for L . We then solve for S . So, the salvage value for the machine is $10,000.

29. Solution b): We want to isolate the variable S , treating all of the other letters in the equation as constants. -C -C This equation allows us to calculate the salvage value for any asset, given the initial cost, estimated life, and depreciation value.