PSME 95 - Final Project

1. PSME 95 - Final Project How to solve for x and y in a system of two equations for a word problem (MATH 210/212) Yu-Jeng (Dennis) Pan Raeda Ashkar

2. Watch Video on Solving Systems by Substitution - Math Help http://www.youtube.com/watch?v=KNwwu5wjcaA

3. Word Problem Solve the word problem for homework Suppose a wealthy American CEO of a financial company earns $1 million from his paycheck each year. His annual net income equals to $1million minus the annual expense. From his past experiences, he believes that his annual expense equals to 25% of his annual net income plus $10000 annual fees on insurance. As the CEO’s accountant, based on the above information, you must determine the CEO’s annual net income and annual expense. But many tutees are afraid of word problems. Some have trouble reading text in the word problem, and some simply have word problem anxieties and refuse to even looking at it.

4. Breaks Word Problem into Small Parts To help tutees to understand what the word problem wants the tutees to do, the tutor can break the word problem into small part. For this example word problem, the tutor can break it into 4 parts: (1) A CEO earns $1 million / year. (2) His annual net income = $1 million minus the annual expense. (3) His annual expense = 25% of his annual net income plus $10000. (4) Find CEO’s annual net income and annual expense. Give tutees 20 seconds to read these 4 parts. After they are done, the tutor should ask tutees to answer a series of questions to solve the world problem step by step – very suitable for tutees who are sequential learners.

5. Step One First variable is “CEO’s annual net income.” Another variable is “CEO’s annual expense.” 1.) What are the 2 variables we need to solve in the word problem? 2.) Substitute the variables in terms of X and Y 3.) The word problem is basically a system of two equations. Please set up the system of two equations. Y = CEO’s annual net income X = CEO’s annual expense Y = 1000000 – X X = 0.25Y + 10000

6. Step Two To solve for Y, plug X = 0.25Y + 10000 into the X variable in Y = 1000000 – X. And got: Y = 1000000 – 0.25Y – 10000 4.) We first solve for Y. But How? 5.) Let us use Algebra to solve for Y 6.) Y is CEO’s annual net income. So CEO’s annual net income is: Y = 1000000 – 0.25Y – 10000 Y = 990000 – 0.25Y 1.25Y = 990000 Y = 990000 / 1.25 Y = 792000 The CEO’s annual net income is Y = $792000

7. Step Three 7.) After knowing Y= 792000, solve for X. But How? 8.) X is CEO’s annual expense. So CEO’s annual expense is: After getting Y = 792000, plug Y = 792000 into X = 0.25Y + 10000 to solve for X: X = 0.25Y + 10000 X = 0.25 (792000) + 10000 X = 198000 + 10000 X = 208000 The CEO’s annual expense is X = $208000

8. Y = 1000000 – X Y = 1000000 – 208000 Y = 792000 X = 0.25Y + 10000 X = 0.25(792000) + 10000 X = 208000 And it works, and so Y = $792000 and X = $208000 is the solution to the system of two equations we need solve. Don’t Forget to Check Work! After finding out that Y = $792000 and X = $208000, plug X and Y into the original system of two equations to check work:

9. Conclusion We are sure that the answer for the word problem is that for the CEO, the annual net income is Y = $792000 and the annual expense is X = $208000.