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Systems of Equations: Substitution
Unit 1, Lesson 1

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

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 ‎

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 ‎

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

Solve by substitution:
teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt ‎

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 ‎

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 ‎

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.

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).

Solve by substitution:
teachers.henrico.k12.va.us/math/.../3-2/3-2SolvSysSubst.ppt ‎

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