1. 1 Topic 5.1.2 The Substitution Method

2. 2 Topic 5.1.2 The Substitution Method California Standard: 9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: You’ll learn about the substitution method and use it to solve systems of linear equations. Key words: substitution system of linear equations simultaneous equations

3. 3 Topic 5.1.2 The Substitution Method In the previous Topic you learned to solve systems of equations by graphing. Graphing can work well if you have integer solutions, but it can be difficult to read off fractional solutions, and even integer solutions if the scale of your graph is small. y = – x + 6 x – y = –4

4. 4 Topic 5.1.2 The Substitution Method Doesn’t Involve Graphs The Substitution Method The substitution method involves a bit more algebra than graphing, but will generally give you more accurate solutions. Substitution Method Substitute this value into one of the earlier equations and solve it to find the value of the second variable. Substitute the expression for the variable into the other equation. This gives an equation with just one variable. Solve this equation to find the value of the variable. Take one of the equations and solve for one of the variables ( x or y ).

5. 5 Topic 5.1.2 Example 1 Solution follows… The Substitution Method Solve this system of equations using substitution: 2 x – 3 y = 7 (Equation 1) –2 x + y = –1 (Equation 2) Solution Step 1: In this case, it’s easiest to solve for y in Equation 2 because it has a coefficient of 1: Rearrange one equation so that one of the variables is expressed in terms of the other. –2 x + y = –1 y = 2 x – 1 (Equation 3) Now you have y expressed in terms of x. Solution continues…

6. 6 Topic 5.1.2 Example 1 The Substitution Method Solve this system of equations using substitution: 2 x – 3 y = 7(Equation 1) –2 x + y = –1(Equation 2) y = 2 x – 1 (Equation 3) Solution Step 2:Substitute 2 x – 1 for y in Equation 1. Then solve to find the value of x. 2 x – 3 y = 7 2 x – 3(2 x – 1) = 7 2 x – 6 x + 3 = 7 –4 x + 3 = 7 –4 x = 4 x = –1 Solution continues…

7. 7 Topic 5.1.2 Example 1 The Substitution Method Substitute –1 for x into an equation to find y. Equation 3 is the best one to use here as y is already isolated — so you don’t have to do any rearranging. y = 2 x – 1 Therefore x = –1, y = –3 is the solution of the system of equations. Solution continues… y = 2(–1) – 1 y = –3 Solve this system of equations using substitution: 2 x – 3 y = 7(Equation 1) –2 x + y = –1 (Equation 2) y = 2 x – 1 (Equation 3) Solution Step 3:

8. 8 Solve this system of equations using substitution: 2 x – 3 y = 7(Equation 1) –2 x + y = –1 (Equation 2) y = 2 x – 1 (Equation 3) Topic 5.1.2 Example 1 The Substitution Method It’s a good idea to check that the solution is correct by substituting it into the original equations. The solution makes both of the original equations true statements, so it must be correct. 2 x – 3 y = 7 2(–1) – 3(–3) = 7 –2 + 9 = 7 7 = 7 — True statement –2 x + y = –1 –2(–1) + (–3) = –1 2 – 3 = –1 –1 = –1 — True statement Solution Step 3: (continued)

9. 9 Topic 5.1.2 Guided Practice Solution follows… The Substitution Method Solve each system of equations in Exercises 1–6 by the substitution method. 1. y = 2 x – 3 and –3 y + 2 x = –15 2. x + 2 y = 7 and 3 x – 2 y = 5 3. a + b = 2 and 5 a – 2 b = –4 –3(2 x – 3) + 2 x = –15 –6 x + 9 + 2 x = –15 –4 x = –24  x = 6, y = 2 × 6 – 3 = 9  x = 6 and y = 9 x = 7 – 2 y  3(7 – 2 y ) – 2 y = 5 21 – 6 y – 2 y = 5 16 = 8 y  y = 2, x = 7 – 2 × 2 = 3  x = 3 and y = 2 a = 2 – b  5(2 – b ) – 2 b = –4 10 – 5 b – 2 b = –4 14 = 7 b  b = 2, a = 2 – 2 = 0  a = 0 and b = 2

10. 10 Topic 5.1.2 Guided Practice Solution follows… The Substitution Method Solve each system of equations in Exercises 1–6 by the substitution method. 4. w + z = 13 and w – 2 z = 4 5. y + 2 x = 5 and 2 x – 3 y = 1 6. m + 6 n = 25 and n – 3 m = –18 y = 5 – 2 x  2 x – 3(5 – 2 x ) = 1 2 x – 15 + 6 x = 1 8 x = 16  x = 2, y = 5 – 2 × 2 = 1 m = 25 – 6 n  n – 3(25 – 6 n ) = –18 n – 75 + 18 n = –18 19 n = 57  n = 3, m = 25 – 6 × 3 = 7 w = 13 – z  13 – z – 2 z = 4 13 – 3 z = 4 9 = 3 z  z = 3, w = 13 – 3 = 10

11. 11 Topic 5.1.2 Independent Practice Solution follows… The Substitution Method Solve by the substitution method: 1. y = 3 x and x + 21 = –2 y 2. x + y = 5 and 5 x + 2 y = 16 3. 7 x – 4 y = 27 and – x + 4 y = 3 4. 9 y – 6 x = 0 and 13 x + 9 = 24 y Solve each system of three equations by the substitution method: 5. 3 x + y = 10, 4 x – z = 7, 7 y + 2 z = 17 6. a + b + c = 7, 2 a + b – 2 c = –11, –4 a – 4 b + c = –32 x = –3, y = –9 x = 2, y = 3 x = 5, y = 2 x = 3, y = 2 x = 3, y = 1, z = 5 a = 2, b = 9, c = 12 1 2 1 3 1 4

12. 12 Topic 5.1.2 The Substitution Method Round Up With the substitution method, it doesn’t matter which equation you choose to rearrange, or which variable you choose to solve for — you will get the same answer as long as you follow the steps correctly and don’t make any mistakes along the way. The important thing is to keep the algebra as simple as you possibly can.