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**Use the substitution method**

EXAMPLE 1 Use the substitution method Solve the linear system: y = 3x + 2 Equation 1 x + 2y = 11 Equation 2 SOLUTION STEP 1 Solve for y. Equation 1 is already solved for y.

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**Use the substitution method**

EXAMPLE 1 Use the substitution method STEP 2 Substitute 3x + 2 for y in Equation 2 and solve for x. x + 2y = 11 Write Equation 2. x + 2(3x + 2) = 11 Substitute 3x + 2 for y. 7x + 4 = 11 Simplify. 7x = 7 Subtract 4 from each side. x = 1 Divide each side by 7.

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EXAMPLE 1 Use the substitution method STEP 3 Substitute 1 for x in the original Equation 1 to find the value of y. y = 3x + 2 = 3(1) + 2 = = 5 ANSWER The solution is (1, 5).

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**Use the substitution method**

EXAMPLE 1 GUIDED PRACTICE Use the substitution method CHECK Substitute 1 for x and 5 for y in each of the original equations. y = 3x + 2 x + 2y = 11 5 = 3(1) + 2 ? 1 + 2 (5) = 11 ? 5 = 5 11 = 11

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**Use the substitution method**

EXAMPLE 2 Use the substitution method Solve the linear system: x – 2y = –6 Equation 1 4x + 6y = 4 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – 2y = –6 Write original Equation 1. x = 2y – 6 Revised Equation 1

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**Use the substitution method**

EXAMPLE 2 Use the substitution method STEP 2 Substitute 2y – 6 for x in Equation 2 and solve for y. 4x + 6y = 4 Write Equation 2. 4(2y – 6) + 6y = 4 Substitute 2y – 6 for x. 8y – y = 4 Distributive property 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14.

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**Use the substitution method**

EXAMPLE 2 Use the substitution method STEP 3 Substitute 2 for y in the revised Equation 1 to find the value of x. x = 2y – 6 Revised Equation 1 x = 2(2) – 6 Substitute 2 for y. x = –2 Simplify. ANSWER The solution is (–2, 2).

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**Use the substitution method**

EXAMPLE 2 GUIDED PRACTICE Use the substitution method CHECK Substitute –2 for x and 2 for y in each of the original equations. Equation 1 Equation 2 4x + 6y = 4 x – 2y = –6 –2 – 2(2) = –6 ? 4(–2) + 6 (2) = 4 ? –6 = –6 4 = 4

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EXAMPLE 1 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. y = 2x + 5 1. 3x + y = 10 ANSWER (1, 7)

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EXAMPLE 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. x – y = 3 2. x + 2y = –6 ANSWER (0, –3)

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EXAMPLE 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. 3x + y = –7 3. –2x + 4y = 0 ANSWER (–2, –1)

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