1. 7.3 S YSTEMS OF L INEAR E QUATIONS IN T WO V ARIABLES

2. O BJECTIVES 1. Decide whether an ordered pair is a solution of a linear system. 2. Solve linear systems by graphing. 3. Solve linear systems by substitution. 4. Solve linear systems by addition. 5. Identify systems that do not have exactly one ordered-pair solution. 6. Solve problems using systems of linear equations.

3. S YSTEM OF L INEAR E QUATIONS IN 2 VARIABLES System of Linear Equations: -- 2 linear equations E.g., 2x – 3y = -4 2x + y = 4 Solution to a System of Equations: -- an ordered pair that satisfy both equations E.g., (1, 2), or, x = 1 and y = 2 is a solution to the system Check 2x – 3y = -4 2x + y = 4 2(1) – 3(2) = -4 2(1) + 2 = 4 2 – 6 = -4 true 2 + 4 = 4 true

4. G RAPHICAL I NTERPRETATION OF THE S OLUTION 2x – 3y = -4 2x + 4 = 3y y = (2/3)x + (4/3) 2x + y = 4 y = -2x + 4 (1, 2)

5. E XAMPLE Solve by graphing the system of equations. x + 2y = 2 an x – 2y = 6 Find x-intercept and y-intercept for both equations. x + 2y = 2 x – 2y = 6 when x = 0, when x = 0, 0 + 2y = 2 0 – 2y = 6 y = 1 y = -3 when y = 0 when y = 0, x + 2(0) = 2 x – 2(0) = 6 x = 2 x = 6 (0, 1) -- y-intercept (0, -3) -- y-intercept (2, 0) -- x-intercept (6, 0) -- x-intercept

6. E XAMPLE ( CONT.) Lines intersect at (4, -1). Check: x + 2y = 2 4 + 2(-1) = 2 4 – 2 = 2 true x – 2y = 6 4 – 2(-1) = 6 4 + 2 = 6 true

7. S OLVING E QUATION S YSTEM BY SUBSTITUTION Solve: y = -x – 1 4x – 3y = 24 Solution Substitute the expression for y in the first equation for y in the second equation. 4x – 3y = 24 y = -x - 1 4x – 3(-x – 1) = 24 y = -(3) - 1 4x + 3x + 3 = 24 y = -4 7x = 21 x = 3 Solution: (3, -4)

8. S OLVING A LINEAR S YSTEM BY S UBSTITUTION Solve y = -x – 1 4x – 3y = 24 This gives us: 4 x – 3(− x – 1) = 24. Solving for x, we get: x = 3 Substitute x value back in the first equation. y = -(3) – 1 This gives us: y = -4 Solution: (3, -4)

9. E XAMPLE Solve the linear system. -4x + y = -11 2x – 3y = 3 Solution 1.Using one equation, express y in terms of x. -4x + y = -11 y = 4x - 11 2.Substitute this in the second equation. 2x – 3y = 3 2x – 3(4x – 11) = 3 3.Solve for x 2x – 12x + 33 = 3 -10x = -30 x = 3

10. E XAMPLE ( CONT.) 4.Substitute this value of x in the first equation and solve for y. y = 4x – 11 y = 4(3) – 11 y = 12 – 11 y = 1 Solution: (3, 1)

11. Y OUR TURN Solve the linear system y = 2x + 7 2x – y = -5 Solution 2x – y = -5 2x – (2x + 7) = -5 2x – 2x + 7 = -5 7 = -5 What does this mean? Check the slopes of the 2 lines.

12. S OLVING A LINEAR SYSTEM BY ADDITION Solve: 3x + 2y = 48 9x – 8y = -24 1.The idea is to eliminate either the x column or the y column and add the two equations. 4(3x + 2y) = 4(48) 9x – 8y = -24 2.12x + 8y = 192 9x – 8y = -24 3.21x = 168 4.x = 8

13. S OLVING A LINEAR SYSTEM BY ADDITION Substitute this value of x in either equation and solve for y. 3x + 2y = 48 3(8) + 2y = 48 24 + 2y = 48 2y = 24 y = 12 Solution: (8, 12)

14. S PECIAL C ASES Number of SolutionsGraphically One ordered-pair solutionTwo lines intersect at one point. No solutionTwo lines are parallel. Infinitely many solutionsTwo lines are the same line.

15. A SYSTEM WITH NO SOLUTION Solve: 4x + 6y = 12 6x + 9y = 12 Using the addition method, 4x + 6y = 12 6x + 9y = 12 0 = 12 false. There is no solution to the system. Multiply by 3 Multiply by -2

16. A SYSTEM WITH INFINITELY MANY SOLUTIONS Solve: y = 3x – 2 15x – 5y = 10 Using the substitution method, 15x – 5y = 10 15x – 5(3x – 2) = 10 15x – 15x + 10 = 10 10 = 10 This is true for any (x, y) pairs. Thus, there is an infinitely number of solutions.

17. M ODELING WITH SYSTEMS OF EQUATIONS Suppose a company produces and sells x iGizmos. Revenue function: R(x) = (price per unit sold)x Cost function: C(x) = fixed cost + (cost per unit produced)x Break-even point: intersection of R(x) and C(x) C(x) = fixed cost + (price per unit produced) x R(x) = fixed cost + (price per unit sold) x Break-even point x (iGizmo units) Dollars

18. F INDING A BREAK - EVEN POINT A company plans to manufacture electronic age wheelchairs. Fixed cost will be $500,000, and the production cost for each wheelchair is $400. The chairs will be sold at $600 apiece. Write the Cost function C(x). Write the Revenue function R(x). Graph the functions. Determine the beak-even point.

19. F INDING A BREAK - EVEN POINT Cost function: C(x) = 500,000 + 400x Revenue function: R(x) = 600x Graph

20. F INDING A BREAK - EVEN POINT Break-even Point C(x) = 500,000 + 400x R(x) = 600x or y = 500,000 + 400x y = 600x 600x = 500,000 + 400x 200x = 500,000 x = 2500

21. B REAK - EVEN POINT x = 2500 y = 600x y = 600(2500) = 1,500,000 Thus, Break-even point: (2500, 1,500,000) I.e., $1,500,000 with 2500 units sold.

22. Y OUR T URN The profit function P(x) = R(x) – C(x) For the preceding case, P(x) = 600x – (500,000 + 400x) P(x) = 200x – 500,000 1.Sketch the graph of the profit function. 2.What is the y-intercept of the function? How do you interpret that value? 3.What is the slope of the function? How do you interpret that value? 4.What is x-intercept of the function? How do you interpret that value?