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**Solving Linear Systems by Substitution**

7.2 Solving Linear Systems by Substitution Objective 1 – Use substitution to solve a linear system. Steps to solve by substitution Isolate a variable in one of the equations. 2. Substitute the expression into the other equation. 3. Solve for one of the variables. 4. Use the variable in step 3 to solve the other variable

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Y = 2x 7x + y = 36 Example Since y = 2x, substitute the y in the second equation with 2x 7x + 2x = 36 9x = 36 x = 4 Since x= 4, Use it to solve for y Y = 2(4) Y = 8 Solution is (4,8)

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Example y = 3 – 3x 7x + 2y = 1 Since y = 3 – 3x, substitute y with 3 – 3x into the 2nd equation 7x + 2(3 – 3x) = 1 7x + 6 – 6x = 1 x + 6 = 1 x = – 5 y = 3 – 3(-5) Since x = – 5, use it to solve for y y = 3 – -15 y = 18 Solution (-5, 18)

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**Solve the linear system**

-x + y = 1 2x + y = -2 EXAMPLE Solve the linear system Solve for y in Equation 1 y = x + 1 Substitute x + 1 for y in equation 2 and solve for x 2x + y = -2 2x + x + 1 = -2 3x + 1 = -2 3x = -3 x = -1 Substitute -1 for x to find y y = x + 1 Solution is ( -1, 0) y = y = 0

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**Solve the linear system**

2x + 2y = 3 x – 4y = -1 EXAMPLE Solve the linear system Solve for x in Equation 2 x = 4y - 1 Substitute 4y - 1 for x in equation 1 and solve for y 2x + 2y = 3 2(4y – 1) + 2y = 3 8y – 2 + 2y = 3 10y = 5 Substitute ½ for y to find x Solution is (1, ½) x = 2 - 1 x = 1

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