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**Systems of Equations: Substitution**

Unit 1, Lesson 1

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Method 2: Substitution Another method for solving a system of equations is called “Substitution”.

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**Steps to Solving: Solve one equation for one variable**

Substitute the expression from step one into the other equation. Simplify and solve the equation. Substitute back into either original equation to find the value of the other variable. teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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**y = 4x 3x + y = -21 Step 1: Solve one equation for one variable.**

Step 2: Substitute into the other equation. 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3 Step 4: Find the value of the other variable. 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12) teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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**Solve using substitution. y = 2x + 2**

–3x + 4 = 2x + 2 Substitute –3x + 4 for y. 4 = 5x + 2 Add 3x to each side. 2 = 5x Subtract 2 from each side. 0.4 = x Divide each side by 5. Solve for the other variable. y = 2(0.4) + 2 Substitute 0.4 for x in either equation. y = Simplify. y = 2.8

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**Solve by substitution:**

teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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**5x -10 + x = 2 6x -10 = 2 6x = 12 x = 2 x + y = 10 y = 10 - x**

Step 1: Solve one equation for one variable. x + y = 10 y = 10 - x Step 2: Substitute into the other equation. 5x -(10 - x) = 2 Step 3: Simplify and solve the equation. 5x x = 2 6x -10 = 2 6x = 12 x = 2 teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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**Solution to the system is (2,8).**

x + y = 10 5x – y = 2 Step 4: Substitute to find the value of the other variable. x + y = 10 2 + y = 10 y = 8 Solution to the system is (2,8). teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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**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 to each side. x = 1.75 Divide each side by 8.

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**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).

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**Solve by substitution:**

teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt

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