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SOLUTION

–6 x – 9 y = –15. EXAMPLE 1. Multiply one equation , then add. Solve the linear system:. 6 x + 5 y = 19. Equation 1. 2 x + 3 y = 5. Equation 2. SOLUTION. STEP 1. Multiply Equation 2 by –3 so that the coefficients of x are opposites . 6 x + 5 y = 19. 6 x + 5 y = 19.

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SOLUTION

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  1. –6x – 9y = –15 EXAMPLE 1 Multiply one equation, then add Solve the linear system: 6x + 5y = 19 Equation 1 2x + 3y = 5 Equation 2 SOLUTION STEP 1 Multiply Equation 2 by –3 so that the coefficients of x are opposites. 6x + 5y = 19 6x + 5y = 19 2x + 3y = 5 STEP 2 Add the equations. –4y = 4

  2. EXAMPLE 1 Multiply one equation, then add STEP 3 Solve for y. y = –1 STEP 4 Substitute –1 for yin either of the original equations and solve for x. 2x + 3y= 5 Write Equation 2. 2x + 3(–1) = 5 Substitute–1 for y. 2x + (–3) = 5 Multiply. Subtract –3 from each side. 2x = 8 x = 4 Divide each side by 2.

  3. ? 2(4)+ 3(–1) = 5 ? 6(4) +5(–1)= 19 19 = 19 5 = 5 EXAMPLE 1 Multiply one equation, then add The solution is (4, –1). ANSWER CHECK Substitute 4 for xand –1 for y in each of the original equations. Equation 2 Equation 1 6x+ 5y= 19 2x+ 3y= 5

  4. EXAMPLE 2 Multiply both equations, then subtract Solve the linear system: 4x + 5y = 35 Equation 1 2y = 3x – 9 Equation 2 SOLUTION STEP 1 Arrange the equations so that like terms are in columns. 4x + 5y = 35 Write Equation 1. –3x + 2y = –9 Rewrite Equation 2.

  5. 8x + 10y = 70 –15x +10y = –45 x = 5 EXAMPLE 2 Multiply both equations, then subtract STEP 2 Multiply Equation 1 by 2 and Equation 2 by 5 so that the coefficient of yin each equation is the least common multiple of 5 and 2, or 10. 4x + 5y = 35 –3x + 2y = –9 Subtract: the equations. STEP 3 23x = 115 Solve: for x. STEP 4

  6. ANSWER The solution is (5, 3). EXAMPLE 2 Multiply both equations, then subtract STEP 5 Substitute 5 for xin either of the original equations and solve for y. 4x+ 5y = 35 Write Equation 1. 4(5)+ 5y = 35 Substitute 5 forx. y = 3 Solve for y.

  7. ? 2(3)= 3(5) – 9 ? 4(5)+ 5(3)= 35 ANSWER The solution is (5, 3). 35= 35 6= 6 EXAMPLE 2 Multiply both equations, then subtract Substitute 5 for xand 3 for yin each of the original equations. CHECK Equation 2 Equation 1 2y= 3x– 9 4x+ 5y= 35

  8. 1. 6x – 2y = 1 ANSWER The solution is (–0.5, –2). for Examples 1 and 2 GUIDED PRACTICE Solve the linear system using elimination. –2x + 3y =–5

  9. 2. 2x + 5y = 3 ANSWER The solution is (9, –3). for Examples 1 and 2 GUIDED PRACTICE Solve the linear system using elimination. 3x + 10y =–3

  10. 3x –7y = 5 3. ANSWER The solution is (–10, –5). for Examples 1 and 2 GUIDED PRACTICE Solve the linear system using elimination. 9y =5x + 5

  11. x + y = 16 x + y = 16 B A x + y = 16 x + y = 76 D C EXAMPLE 3 Standardized Test Practice Darlene is making a quilt that has alternating stripes of regular quilting fabric and sateen fabric. She spends $76 on a total of 16 yards of the two fabrics at a fabric store. Which system of equations can be used to find the amount x(in yards) of regular quilting fabric and the amount y(in yards) of sateen fabric she purchased? 4x + 6y = 76 x + y = 76 4x + 6y = 16 6x + 4y = 76

  12. x 16 y + = EXAMPLE 3 Standardized Test Practice SOLUTION Write a system of equations where xis the number of yards of regular quilting fabric purchased and yis the number of yards of sateen fabric purchased. Equation1: Amount of fabric

  13. x y + 6 4 = 76 ANSWER The correct answer is B. B C D A EXAMPLE 3 Standardized Test Practice Equation2:Cost of fabric The system of equations is: x + y = 16 Equation 1 4x + 6y = 76 Equation 2

  14. SOCCER A sports equipment store is having a sale on soccer balls. A soccer coach purchases 10 soccer balls and 2 soccer ball bags for $155. Another soccer coach purchases 12 soccer balls and 3 soccer ball bags for $189. Find the cost of a soccer ball and the cost of a soccer ball bag. 4. ANSWER soccer ball $14.50, soccer ball bag: $5 for Example 3 GUIDED PRACTICE

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