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**3.2 Solving Systems Algebraically**

Substitution & Elimination

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**Equation 1 Equation 2 Revised Equation 2 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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**Write revised Equation 2.**

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( ) + 5y = –5 –3y + 3 Substitute –3y + 3 for x. y = 11 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( ) + 3 Substitute 11 for y. x = –30 Simplify.

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**Substitute for x and y. Solution checks. 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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**Equation 1 Equation 2 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. –6x + 14y = -20 3x – 7y = 10 6x – 8y = 8 6x – 8y = 8 STEP 2 6y = –12 Add the revised equations and solve for y. y = –2

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**Write Equation 1. Substitute –2 for y. Simplify. Solve for x.**

EXAMPLE 2 Use the elimination method 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 for 1 and the elimination method for 2. 1. 4x + 3y = –2 x + 5y = –9 2. 3x + 3y = –15 5x – 9y = 3 The solution is (1,–2). ANSWER The solution is ( -3 , –2) 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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**Write first equation. Solve for x. 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. Because the coefficient of x in the first equation is 1, use the substitution method. Solve the first equation for x. x – 2y = 4 Write first equation. x = 2y + 4 Solve for x.

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**Write second equation. Substitute 2y + 4 for x. Simplify. EXAMPLE 4**

Solve linear systems with many or no solutions Substitute the expression for x into the second equation. 3x – 6y = 8 Write second equation. 3(2y + 4) – 6y = 8 Substitute 2y + 4 for x. 12 = 8 Simplify. Because the statement 12 = 8 is never true, there is no solution. ANSWER

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EXAMPLE 4 Solve linear systems with many or no solutions b. Because no coefficient is 1 or – 1, use the elimination method. Multiply the first equation by 7 and the second equation by 2. 4x – 10y = 8 28x – 70y = 56 – 14x + 35y = – 28 – 28x + 70y = – 56 Add the revised equations. 0 = 0 ANSWER Because the equation 0 = 0 is always true, there are infinitely many solutions.

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**Write second equation. Solve for y. GUIDED PRACTICE for Example 4**

Solve the linear system using any algebraic method. 5. 12x – 3y = – 9 – 4x + y = 3 SOLUTION Because the coefficient of y in the second equation is y, use the substitution method. Solve the 2nd equation for y. – 4x + y = 3 Write second equation. y = 4x + 3 Solve for y.

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**Write second equation. Substitute 4x + 3 for y. Simplify.**

GUIDED PRACTICE for Example 4 Substitute the expression for y into the first equation. 12x – 3y = – 9 Write second equation. 12x – 3(4y + 3) = – 9 Substitute 4x + 3 for y. 0 = 0 Simplify. Because the equation 0 = 0 is always true, there are infinitely many solutions. ANSWER

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GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 6. 6x + 15y = – 12 – 2x – 5y = 9 Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 3 6x + 15y = – 12 6x + 15y = – 12 – 2x – 5y = 9 3 – 6x – 15y = 27 0 = 15 Add the revised equations. ANSWER Because the statement 0 = 15 is never true, there are no solutions.

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**Write second equation. Solve for x. GUIDED PRACTICE for Example 4**

Solve the linear system using any algebraic method. 7. 5x + 3y = 20 – x – y = – 4 3 5 Because the coefficient of x in the first equation is – 1, use the substitution method. Solve the 2nd equation for x. – x – y = – 4 3 5 Write second equation. x = – y + 4 3 5 Solve for x.

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**Write first equation. Substitute for x. Simplify. GUIDED PRACTICE**

for Example 4 Substitute the expression for x into the first equation. 5x + 3y = 20 Write first equation. 5 ( – y + 4 ) + 3y = 20 3 Substitute for x. 5 3 – y + 4 20 = 20 Simplify. Because the statement 20 = 20 is always true, there are infinitely many solution. ANSWER

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GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 8. 12x – 2y = 21 3x + 12y = – 4 Because no coefficient is 1 or – 1, use the elimination method. Multiply the second equation by 6 12x – 2y = 21 6 72x – 12y = 126 3x + 12y = – 4 3x + 12y = – 4 Add the revised equations. 75x = 122 x 122 75 =

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**Write second equation. 122 Substitute for x. 75 Simplify.**

GUIDED PRACTICE for Example 4 Substitute the expression for x into the second equation. 3x + 12y = – 4 Write second equation. 3( ) + 12y = 8 122 75 Substitute for x. 122 75 y = – 37 50 Simplify. The solution is ( , ) ANSWER 122 75 – 37 50

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GUIDED PRACTICE for Example 4 Solve the linear system using any algebraic method. 5x + 5y = 5 10. 5x + 3y = 4.2 9. 8x + 9y = 15 5x – 2y = 17 (3, –1) ANSWER (0.6, 0.4) ANSWER

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