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Solve Linear Systems by Substitution January 28, 2014 Pages 435-438

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7.2 SOLVING SYSTEM BY SUBSTITUTION Students will be able to: Solve linear equations in two variables by substitution.

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

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

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

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CHECK y = 3x + 2 5 = 3(1) + 2 ? 5 = 5 Substitute 1 for x and 5 for y in each of the original equations. x + 2y = 11 1 + 2 ( 5 ) = 11 ? 11 = 11

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

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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. Distributive property 8y – 24 + 6y = 4 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14. STEP 2

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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). STEP 3

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

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3. Solve the linear system using the substitution method. Equation 2 Equation 1 3x + y = 10 Solve for y. Equation 1 is already solved for y. SOLUTION STEP 1 y = 2x + 5

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Substitute 2x + 5 for y in Equation 2 and solve for x. 3x + y = 10 Write Equation 2. 3x + (2x + 5) = 10 Substitute 2x+5 for x. 5x + 5 = 10 Simplify. 5x = 5 x = 1 STEP 2

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Substitute 1 for x in the revised Equation 1 to find the value of y. ANSWER The solution is ( 1, 7). y = 2x + 5 = 2(1) + 5 = 7 STEP 3

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Equation 1 x + 2y = – 6 Equation 2 SOLUTION Solve Equation 1 for x. x – y = 3 Write original Equation 1. x = y + 3 Revised Equation 1 STEP 1 x – y = 3 4. Solve the linear system using the substitution method.

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Substitute y + 3 for x in Equation 2 and solve for y. x + 2y = – 6 Write Equation 2. ( y + 3) + 2y = – 6 Substitute y + 3 for x. 3y + 3 = – 6 Simplify. y = – 3 Divide each side by 3. STEP 2 3y = – 9 Add 3 to each side.

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Substitute – 3 for y in the revised Equation 1 to find the value of x. x = y + 3 Revised Equation 1 x = – 3 + 3 Substitute – 3 for y. x = 0 Simplify. ANSWER The solution is ( 0, – 3 ). STEP 3

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Equation 1 –2x + 4y = 0 Equation 2 SOLUTION Solve Equation 2 for x. -2x + 4y = 0 Write original Equation 2. x = 2y Revised Equation 2 STEP 1 3x + y = –7 5. Solve the linear system using the substitution method.

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Substitute 2y for x in Equation 1 and solve for y. 3x + y = – 7 Write Equation 2. 3( 2y ) + y = – 7 Substitute 2 y for x. 6y + y = – 7 Distributive property. y = – 1 Divide each side by 7. STEP 2 7y = – 7 Simplify.

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

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HOMEWORK Pages 439, Problems #4-16, even, #20-24, even

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