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**Use the substitution method**

EXAMPLE 1 Use the substitution method Solve the system using the substitution method. 2x + 5y = –5 Equation 1 x + 3y = 3 Equation 2 SOLUTION STEP 1 Solve Equation 2 for x. x = –3y + 3 Revised Equation 2

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**Use the substitution method**

EXAMPLE 1 Use the substitution method STEP 2 Substitute the expression for x into Equation 1 and solve for y. 2x +5y = –5 Write Equation 1. 2(–3y + 3) + 5y = –5 Substitute –3y + 3 for x. y = 11 Solve for y. STEP 3 Substitute the value of y into revised Equation 2 and solve for x. x = –3y + 3 Write revised Equation 2. x = –3(11) + 3 Substitute 11 for y. x = –30 Simplify.

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**Use the substitution method**

EXAMPLE 1 Use the substitution method The solution is (– 30, 11). ANSWER CHECK Check the solution by substituting into the original equations. 2(–30) + 5(11) –5 = ? Substitute for x and y. = ? –30 + 3(11) 3 –5 = –5 Solution checks. 3 = 3

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**Use the elimination method**

EXAMPLE 2 Use the elimination method Solve the system using the elimination method. 3x – 7y = 10 Equation 1 6x – 8y = 8 Equation 2 SOLUTION STEP 1 Multiply Equation 1 by – 2 so that the coefficients of x differ only in sign. 3x – 7y = 10 –6x + 14y = 220 6x – 8y = 8 6x – 8y = 8

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**Use the elimination method**

EXAMPLE 2 Use the elimination method STEP 2 Add the revised equations and solve for y. 6y = –12 y = –2 STEP 3 Substitute the value of y into one of the original equations. Solve for x. 3x – 7y = 10 Write Equation 1. 3x – 7(–2) = 10 Substitute –2 for y. 3x + 14 = 10 Simplify. x = 4 3 – Solve for x.

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EXAMPLE 2 Use the elimination method The solution is ( , –2) 4 3 – ANSWER CHECK You can check the solution algebraically using the method shown in Example 1. You can also use a graphing calculator to check the solution.

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GUIDED PRACTICE for Examples 1 and 2 Solve the system using the substitution or the elimination method. 1. 4x + 3y = –2 x + 5y = –9 2. 3x + 3y = –15 5x – 9y = 3 The solution is (1,–2). ANSWER The solution is ( , –2) 3 – ANSWER

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GUIDED PRACTICE for Examples 1 and 2 Solve the system using the substitution or the elimination method. 3. 3x – 6y = 9 –4x + 7y = –16 ANSWER The solution is (11, 4)

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EXAMPLE 4 Solve linear systems with many or no solutions Solve the linear system. a.x – 2y = 4 3x – 6y = 8 b.4x – 10y = 8 – 14x + 35y = – 28 SOLUTION a.

EXAMPLE 4 Solve linear systems with many or no solutions Solve the linear system. a.x – 2y = 4 3x – 6y = 8 b.4x – 10y = 8 – 14x + 35y = – 28 SOLUTION a.

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