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Solve a System Graphically |

Learn how to solve a system of linear equations by graphing them and estimating the solution |

jmercer
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Solve a System Graphically |

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  1. 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. EXAMPLE 1 Solve a system graphically Graph the linear system and estimate the solution. Then check the solution algebraically. 4x + y = 8 Equation 1 2x – 3y = 18 Equation 2 SOLUTION

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

  3. 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. 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 Equation 1 8x – 6y = 16 Equation 2 SOLUTION

  4. 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. EXAMPLE 3 Solve a system with no solution Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. Equation 1 2x + y = 4 Equation 2 2x + y = 1 SOLUTION

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

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

  7. 50= 20+ 30 50= 2.5(20) The total costs are equal after 20 rides. ANSWER The correct answer is B. EXAMPLE 4 Standardized Test Practice Notice that you need to graph the equations only in the first quadrant because only nonnegative values of xand ymake 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.

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

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

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

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

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

  13. 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. 4 3 The solution is ( , – 2) – EXAMPLE 2 Use the elimination method

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

  15. EXAMPLE 3 Standardized Test Practice Equation 1 Equation 2

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

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

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

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

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

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