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Lesson 7.2- Substitution , pg. 376

Lesson 7.2- Substitution , pg. 376. Goal 1: To solve systems of equations by using substitution. Goal 2: To solve real-world problems involving systems of equations. Vocabulary. Substitution- an algebraic method used to find the exact solution of a linear system. Substitution Method.

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Lesson 7.2- Substitution , pg. 376

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  1. Lesson 7.2- Substitution, pg. 376 • Goal 1: To solve systems of equations by using substitution. • Goal 2: To solve real-world problems involving systems of equations.

  2. Vocabulary • Substitution- an algebraic method used to find the exact solution of a linear system.

  3. Substitution Method • Solve one of the equations for one of it’s variables.(Hint: Solve the equation that contains a variable with leading coefficient of one). • Substitute the expression found in Step 1 intothe remaining equation. • Solve for the remaining variable. • Substitute the value of the variable found in Step 3 into either original equation then solve.

  4. Example 1: Solve using substitution. • x = 4y 4x – y = 75 • Substitute 4y in place of x in Eq. 2 4(4y) – y = 75 Eq. 2 16y – y = 75 15y = 75 y = 5 • Substitute 5 for y in either equation to find the value of x. x = 4(5) x = 20 Solution (20, 5)

  5. 4x + y = 12 Eq. 1 -2x – 3y = 14 Eq. 2 Solve Eq. 1 for y. 4x + y = 12 Eq. 1 y = -4x + 12 Substitute -4x + 12 in place of y in Eq. 2 -2x – 3( -4x + 12) = 14Eq. 2 -2x + 12x – 36 = 14 Distributive Prop. 10x – 36 = 14 Combine Like Terms 10x = 50Add 36 x = 5 c) Substitute 5 in place of x in either equation to find y. y = -4x + 12 Revised equation y = -4(5) + 12 y = -20 + 12 y = -8 Solution (5, -8)

  6. 2x + 2y = 8 Eq. 1 x + y = -2 Eq. 2 a) Solve Eq. 2 for y. x + y = -2 Eq. 2 y = -x – 2 b) Substitute –x – 2 in place of y in Eq. 1 2x + 2(-x – 2) = 8 2x – 2x – 4 = 8 distributive property -4 ≠ 8 No solution

  7. 6x – 2y = -4 Eq. 1 y = 3x + 2 Eq. 2 • Substitute 3x + 2 in place of y in Eq. 1 6x – 2(3x + 2) = -4 6x – 6x – 4 = -4 distributive property -4 = -4combine like terms Infinite Solutions

  8. y = 3x – 8 Eq. 1 y = 4 – x Eq. 2 • Substitute 3x – 8 in place of y in Eq. 2 3x – 8 = 4 – x 4x – 8 = 4 Add x 4x = 12 Add 8 x = 3Divide by 4 • Substitute 3 in place of x into either equation to solve for y. y = 3x – 8 Eq. 1 y = 3(3) – 8 Replace x with 3 y = 1 Solution (3, 1)

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