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3.2 Solving Systems Algebraically

3.2 Solving Systems Algebraically. Substitution & Elimination. Solve Equation 2 for x. STEP 1. EXAMPLE 1. Use the substitution method. Solve the system using the substitution method. 2 x + 5 y = –5. Equation 1. x + 3 y = 3. Equation 2. SOLUTION. x = –3 y + 3.

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3.2 Solving Systems Algebraically

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  1. 3.2 Solving Systems Algebraically Substitution & Elimination

  2. 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

  3. 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( ) + 5y = –5 –3y + 3 Substitute –3y + 3 for x. y = 11 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() + 3 Substitute 11 for y. x = –30 Simplify.

  4. 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.

  5. 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 STEP 2 6y = –12 Add the revised equations and solve for y. y= –2

  6. 4 – x= 3 EXAMPLE 2 Use the elimination method 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.

  7. 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

  8. ANSWER 1. 4x + 3y = –2 The solution is (1,–2). x + 5y = –9 ANSWER 3x + 3y = –15 2. 5x – 9y = 3 The solution is (-3 , –2) for Examples 1 and 2 GUIDED PRACTICE Solve the system using the substitution for 1 and the elimination method for 2.

  9. 3x – 6y = 9 3. –4x + 7y = –16 for Examples 1 and 2 GUIDED PRACTICE Solve the system using the substitution or the elimination method. ANSWER The solution is (11, 4)

  10. 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.

  11. 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.

  12. 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.

  13. 5. 12x – 3y = – 9 – 4x + y = 3 for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method. SOLUTION Because the coefficient of yin the second equation is y, use the substitution method. Solve the 2nd equation for y. – 4x + y = 3 Write second equation. y = 4x + 3 Solve for y.

  14. ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions. for Example 4 GUIDED PRACTICE Substitute the expression for yinto the first equation. 12x – 3y = – 9 Write second equation. 12x – 3(4y + 3)= – 9 Substitute 4x + 3 for y. 0 = 0 Simplify.

  15. 6. 6x + 15y = – 12 – 2x – 5y = 9 3 0 = 15 ANSWER Because the statement 0 = 15 is never true, there are no solutions. for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method. Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 3 6x + 15y = – 12 6x + 15y = – 12 – 6x – 15y = 27 – 2x – 5y = 9 Add the revised equations.

  16. 7. 5x + 3y = 20 – x –y = – 4 – x –y = – 4 3 3 3 x = – y + 4 5 5 5 for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method. Because the coefficient of xin the first equation is –1, use the substitution method. Solve the 2nd equation for x. Write second equation. Solve for x.

  17. ( – y + 4 ) + 3y = 20 5 Substitute for x. – y + 4 ANSWER 3 5 Because the statement 20 = 20 is always true, there are infinitely many solution. 3 5 for Example 4 GUIDED PRACTICE Substitute the expression for xinto the first equation. Write first equation. 5x + 3y = 20 20 = 20 Simplify.

  18. 8. 12x – 2y = 21 3x + 12y = – 4 6 122 75 x = for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method. Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 6 12x – 2y = 21 72x – 12y = 126 3x + 12y = – 4 3x + 12y = – 4 Add the revised equations. 75x = 122

  19. 122 3( ) + 12y = 8 Substitute for x. 75 y = ANSWER The solution is ( , ) – 37 – 37 122 122 75 50 50 75 for Example 4 GUIDED PRACTICE Substitute the expression for xinto the second equation. 3x + 12y = – 4 Write second equation. Simplify.

  20. 10. 5x + 5y = 5 9. 8x + 9y = 15 5x + 3y = 4.2 5x – 2y = 17 ANSWER ANSWER (3, –1) (0.6, 0.4) for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method.

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