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Solving Linear Systems: Substitution vs. Elimination

Learn to solve linear systems with the substitution and elimination methods. Explore examples with different solution possibilities - infinitely many solutions or no solutions.

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Solving Linear Systems: Substitution vs. Elimination

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  1. 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.

  2. 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.

  3. 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.

  4. 6. 6x + 15y = – 12 5. 12x – 3y = – 9 – 2x – 5y = 9 –4x + y = 3 ANSWER infinitely many solutions ANSWER no solutions for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method.

  5. 8. 12x – 2y = 21 7. 5x + 3y = 20 3x + 12y = – 4 ANSWER ANSWER – x –y = – 4 infinitely many solutions The solution is ( , ) 3 122 – 37 75 50 5 for Example 4 GUIDED PRACTICE Solve the linear system using any algebraic method.

  6. 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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