6 2 Solving Systems by Substitution Lesson Objective

6 -2 Solving Systems by Substitution Lesson Objective: I will be able to … • Solve systems of linear equations in two variables by substitution Language Objective: I will be able to … • Read, write, and listen about vocabulary, key concepts, and examples Holt Algebra 1

6 -2 Solving Systems by Substitution Page 5 Solving Systems of Equations by Substitution Step 1 Solve for one variable in at least one equation, if necessary. Step 2 Substitute the resulting expression into the other equation. Step 3 Solve that equation to get the value of the first variable. Step 4 Step 5 Substitute that value into one of the original equations and solve. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check. Holt Algebra 1

6 -2 Solving Systems by Substitution Page 6 Example 1: Solving a System of Linear Equations by Substitution Solve the system by substitution. y = 3 x y=x– 2 Step 1 y = 3 x y=x– 2 Step 2 y= x– 2 3 x = x – 2 Step 3 –x –x 2 x = – 2 2 2 x = – 1 Holt Algebra 1 Step 4 Step 5 y = 3 x y = 3(– 1) y = – 3 (– 1, – 3) Check Substitute (– 1, – 3) into both equations in the system. y=x– 2 y = 3 x – 3 3(– 1) – 3 – 1 – 2 – 3 – 3

6 -2 Solving Systems by Substitution Example 2: Solving a System of Linear Page 7 Equations by Substitution Solve the system by substitution. Step 4 y = x + 1 y=x+1 y=1+1 4 x + y = 6 y=2 Step 1 y = x + 1 Step 5 (1, 2) Step 2 4 x + y = 6 Check Substitute (1, 2) 4 x + (x + 1) = 6 into both equations in the 5 x + 1 = 6 system. Step 3 – 1 y=x+1 4 x + y = 6 5 x = 5 2 1+1 4(1) + 2 6 5 5 2 2 6 6 x=1 Holt Algebra 1

6 -2 Solving Systems by Substitution Page 8 Your Turn 2 Solve the system by substitution. x = 2 y – 4 x + 8 y = 16 Step 4 x + 8 y = 16 x + 8(2) = 16 Step 1 x = 2 y – 4 x + 16 = 16 – 16 Step 2 x + 8 y = 16 x = 0 (2 y – 4) + 8 y = 16 Step 5 (0, 2) Step 3 10 y – 4 = 16 Check Substitute (0, 2) into +4 +4 10 y = 20 both equations in the system. x + 8 y = 16 10 10 x = 2 y – 4 y=2 0 2(2) – 4 0 + 8(2) 16 16 16 0 0 Holt Algebra 1

6 -2 Solving Systems by Substitution Page 9 Solve Example 3: Using the Distributive Property y + 6 x = 11 3 x + 2 y = – 5 by substitution. Step 1 y + 6 x = 11 – 6 x y = – 6 x + 11 Step 2 3 x + 2 y = – 5 3 x + 2(– 6 x + 11) = – 5 Step 4 y + 6 x = 11 y + 6(3) = 11 y + 18 = 11 – 18 y = – 7 Holt Algebra 1 Step 3 3 x + 2(– 6 x) + 2(11) = – 5 3 x – 12 x + 22 = – 5 – 9 x + 22 = – 5 – 22 – 9 x = – 27 – 9 x=3 Step 5 (3, – 7)

6 -2 Solving Systems by Substitution Caution When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. Holt Algebra 1

6 -2 Solving Systems by Substitution Page 10 Solve Your Turn 3 – 2 x + y = 8 3 x + 2 y = 9 Step 1 – 2 x + y = 8 + 2 x +2 x y = 2 x + 8 Step 2 3 x + 2 y = 9 3 x + 2(2 x + 8) = 9 by substitution. Step 3 3 x + 2(2 x) + 2(8) = 9 3 x + 4 x + 16 = 9 7 x + 16 = 9 – 16 7 x = – 7 Step 4 – 2 x + y = 8 – 2(– 1) + y = 8 y+2=8 – 2 y Holt Algebra 1 =6 7 7 x = – 1 Step 5 (– 1, 6)

6 -2 Solving Systems by Substitution Classwork Assignment #31 • 6 -2 Practice B / 6 -1 Practice C Worksheet Holt Algebra 1

6 -2 Solving Systems by Substitution Homework Assignment #31 • Finish 6 -2 / 6 -1 Worksheet • Holt 6 -2 #8 – 11, 15, 17, 24, 25, 36, 37 • KIN 6 -3 Holt Algebra 1