6-2 Solving Systems Using Substitution Hubarth Algebra.

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6-2 Solving Systems Using Substitution Hubarth Algebra

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

Ex 2 Using Substitution and the Distributive Property Step 1:Solve the first equation for x because it has a coefficient of -1. Step 2:Write an equation containing only one variable and solve. y = 3 Divide each side by 6. Step 3:Solve for y in the other equation. -x + 2(3) = 8 Substitute 3 for y. -x + 6 =8 Simplify and divide each side by -1. x = -2 Subtract 2 from each side. (-2, 3)

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

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