6-2 Solving Systems Using Substitution

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Presentation transcript:

6-2 Solving Systems Using Substitution Hubarth Algebra

Ex 1 Solve Systems by Substitution Solve using substitution. y = 2x + 2 y = –3x + 4 Step 1: Write an equation containing only one variable and solve. y = –3x + 4 y = 2x + 2 –3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation. 4 = 5x + 2 Add 3x to each side. 2 = 5x Subtract 2 from each side. 0.4 = x Divide each side by 5. Step 2: Solve for the other variable. y = 2(0.4) + 2 Substitute 0.4 for x in either equation. y = 0.8 + 2 Simplify. y = 2.8 The solution is ( 0.4, 2.8)

Ex 2 Using Substitution and the Distributive Property Solve using substitution. –2x + y = –1 4x + 2y = 12 Step 1: Solve the first equation for y because it has a coefficient of 1. –2x + y = –1 y = 2x –1 Add 2x to each side. Step 2: Write an equation containing only one variable and solve. 4x + 2y = 12 Start with the other equation. 4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation. 4x + 4x – 2 = 12 Use the Distributive Property. 8x = 14 Combine like terms and add 2 to each side. x = 1.75 Divide each side by 8. Step 3: Solve for y in the other equation. –2(1.75) + y = –1 Substitute 1.75 for x. –3.5 + y = –1 Simplify. y = 2.5 Add 3.5 to each side. The solution is (1.75, 2.5)

Ex 3 Solve System by Using Substitution y = 4x – 1 5 = 6x – y ( ) ( ) 5 = 6x – (4x – 1) y = 4 (2) – 1 5 = 6x – 4x + 1 y = 8 – 1 y = 7 5 = 2x +1 4 = 2x x = 2 (2, 7) is the solution of the system

Practice y = x – 3 x + y = 5 2. 2x + y = 1 x = -2y + 5 (4, 1) (-1, 3) 4. x = 4y – 1 3x + 5y = 31 3. x – 5 = y 2x – 3y = 7 (7, 2) (8, 3)