1. EXAMPLE 1 Solve a system graphically Graph the linear system and estimate the solution. Then check the solution algebraically. 4x + y = 8 2x – 3y = 18 Equation 1 Equation 2 SOLUTION Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at ( 3, – 4 ). You can check this algebraically as follows.

2. EXAMPLE 1 Solve a system graphically Equation 1 Equation 2 4x + y = 8 4 ( 3 ) + ( – 4 ) 8 = ? = ? 12 – 4 8 8 = 8 2x – 3y = 18 = ? 2(3) – 3( – 4) 18 = ? 6 + 12 18 18 = 18 The solution is ( 3, – 4 ).

3. EXAMPLE 2 Solve a system with many solutions Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. 4x – 3y = 8 8x – 6y = 16 Equation 1 Equation 2 SOLUTION The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent.

4. EXAMPLE 3 Solve a system with no solution Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. 2x + y = 4 2x + y = 1 Equation 1 Equation 2 SOLUTION The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent.

5. EXAMPLE 4 Standardized Test Practice SOLUTION Equation 1 (Option A) y = 1 x + 30

6. Equation 2 (Option B) EXAMPLE 4 Standardized Test Practice y= x 2.5 To solve the system, graph the equations y = x + 30 and y = 2.5x, as shown at the right.

7. EXAMPLE 4 Standardized Test Practice Notice that you need to graph the equations only in the first quadrant because only nonnegative values of x and y make sense in this situation. The lines appear to intersect at about the point ( 20, 50 ). You can check this algebraically as follows. Equation 1 checks. Equation 2 checks. 50 = 20 + 30 50 = 2.5(20) ANSWER The total costs are equal after 20 rides. The correct answer is B.

8. EXAMPLE 1 Use the substitution method Solve the system using the substitution method. 2x + 5y = – 5 x + 3y = 3 Equation 1 Equation 2 SOLUTION STEP 1 Solve Equation 2 for x. x = – 3y + 3 Revised Equation 2

9. EXAMPLE 1 Use the substitution method STEP 2 Substitute the expression for x into Equation 1 and solve for y. 2x +5y = – 5 2(– 3y + 3) + 5y = – 5 y = 11 Write Equation 1. Substitute – 3y + 3 for x. Solve for y. STEP 3 Substitute the value of y into revised Equation 2 and solve for x. x = – 3y + 3 x = – 3(11) + 3 x = – 30 Write revised Equation 2. Substitute 11 for y. Simplify.

10. EXAMPLE 1 Use the substitution method CHECK Check the solution by substituting into the original equations. 2(– 30) + 5(11) – 5 = ? Substitute for x and y. = ? – 30 + 3(11) 3 Solution checks. 3 = 3– 5 = – 5 The solution is ( – 30, 11 ). ANSWER

11. EXAMPLE 2 Use the elimination method Solve the system using the elimination method. 3x – 7y = 10 6x – 8y = 8 Equation 1 Equation 2 SOLUTION Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign. STEP 1 3x – 7y = 10 6x – 8y = 8 – 6x + 14y = 20 6x – 8y = 8

12. EXAMPLE 2 Use the elimination method STEP 2 Add the revised equations and solve for y. 6y = – 12 y = – 2 STEP 3 Substitute the value of y into one of the original equations. Solve for x. 3x – 7y = 10 3x – 7( – 2) = 10 3x + 14 = 10 x =x = 4 3 – Solve for x. Simplify. Substitute – 2 for y. Write Equation 1.

13. EXAMPLE 2 Use the elimination method The solution is (, – 2 ) 4 3 – ANSWER CHECK You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

14. EXAMPLE 3 Standardized Test Practice SOLUTION Write verbal models for this situation.

15. Equation 1 EXAMPLE 3 Standardized Test Practice Equation 2

16. EXAMPLE 3 Standardized Test Practice STEP 2 Write a system of equations. Equation 1 Equation 2 5x + 7y = 2500 8x + 12y = 4200 Total cost for all T-shirts Total revenue from all T-shirts sold STEP 3 Solve the system using the elimination method. Multiply Equation 1 by – 8 and Equation 2 by 5 so that the coefficients of x differ only in sign. 5x + 7y = 2500 8x + 12y = 4200 – 40x – 56y = – 20,000 40x + 60y = 21,000 Add the revised equations and solve for y. 4y = 1000 y = 250

17. EXAMPLE 3 Standardized Test Practice Substitute the value of y into one of the original equations and solve for x. 5x + 7y = 2500 5x + 7(250) = 2500 5x + 1750 = 2500 x = 150 Write Equation 1. Substitute 250 for y. Simplify. Solve for x. The school sold 150 short sleeve T-shirts and 250 long sleeve T-shirts. ANSWER The correct answer is C.

18. EXAMPLE 4 Solve linear systems with many or no solutions Solve the linear system. a.x – 2y = 4 3x – 6y = 8 b.4x – 10y = 8 – 14x + 35y = – 28 SOLUTION a. Because the coefficient of x in the first equation is 1, use the substitution method. Solve the first equation for x. x – 2y = 4 x = 2y + 4 Write first equation. Solve for x.

19. EXAMPLE 4 Solve linear systems with many or no solutions Substitute the expression for x into the second equation. 3x – 6y = 8 3(2y + 4) – 6y = 8 12 = 8 Write second equation. Substitute 2y + 4 for x. Simplify. Because the statement 12 = 8 is never true, there is no solution. ANSWER

20. EXAMPLE 4 Solve linear systems with many or no solutions b. Because no coefficient is 1 or – 1, use the elimination method. Multiply the first equation by 7 and the second equation by 2. 4x – 10y = 8 – 14x + 35y = – 28 28x – 70y = 56 – 28x + 70y = – 56 Add the revised equations. 0 = 0 ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions.