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**Solving Systems by Substitution**

Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

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Warm Up Solve the linear equations for x. 1. y = x + 3 2. y = 3x – 4 Simplify each expression. x = y – 3 3. 2(x – 5) 2x – 10 4. 12 – 3(x + 1) 9 – 3x

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Warm Up Continued Evaluate each expression for the given value of x. 5. x + 8 for x = 6 6. 3(x – 7) for x =10 12 9

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Objective Solve systems of linear equations in two variables by substitution.

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Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2.

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**Solving Systems of Equations by Substitution**

Step 2 Step 3 Step 4 Step 5 Pick one equation and solve for one variable. Circle this equation. Step 1 Substitute the resulting expression into the other equation. Solve that equation to get the value of the first variable. Substitute this value back into the circled equation from Step 1. Write the values from steps 3 and 4 as an ordered pair, (x, y), and check.

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Example 1: Solve the system by substitution. y = x + 1 4x + y = 6

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Example 2: Solve the system by substitution. x + 2y = –1 x – y = 5

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Example 3: Solve the system by substitution. y – x = 3 y = 2x + 5

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Example 4: Solve the system by substitution. x = 2y – 4 x + 8y = 16

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Example 5: Solve the system by substitution. 2x + y = –4 x + y = –7

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Example 6: y + 6x = 11 Solve by substitution. 3x + 2y = –5

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Example 7: –2x + y = 8 Solve by substitution. 3x + 2y = 9

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Lesson Quiz: Part I Solve each system by substitution. 1. 2. 3. y = 2x (–2, –4) x = 6y – 11 (1, 2) 3x – 2y = –1 –3x + y = –1 x – y = 4

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