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U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS

U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS. 1. 2. 3. THE SUBSTITUTION METHOD. In this lesson you will study two algebraic methods for solving linear systems. The first method is called substitution. Solve one of the equations for one of its variables.

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U SING A LGEBRAIC M ETHODS TO S OLVE S YSTEMS

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  1. USING ALGEBRAIC METHODS TO SOLVE SYSTEMS 1 2 3 THE SUBSTITUTION METHOD In this lesson you will study two algebraic methods for solving linear systems. The first method is called substitution. Solve one of the equations for one of its variables. Substitute expression from Step 1 into other equation and solve for other variable. Substitute value from Step 2 into revised equation from Step 1. Solve.

  2. The Substitution Method 3x + 4y  – 4Equation 1 x + 2y  2Equation 2 Solve the linear system using the substitution method. SOLUTION Solve Equation 2 forx. x + 2y  2 Write Equation 2. x –2y + 2 Revised Equation 2. Substitute the expression forxinto Equation 1 and solve for y. 3x + 4y  – 4 Write Equation 1. Substitute – 2y + 2for x. 3(–2y + 2) + 4y  – 4 y5 Simplify.

  3. The Substitution Method Substitute the expression forxinto Equation 1 and solve fory. 3x + 4y  – 4 Write Equation 1. 3(–2y + 2) + 4y  – 4 Substitute –2y + 2for x. y5 Simplify. 3x + 4y  – 4Equation 1 x + 2y  2Equation 2 Solve the linear system using the substitution method. Substitute the value ofyinto revised Equation 2 and solve forx. x –2y+ 2 Write revised Equation 2. Substitute 5for y. x –2(5) + 2 x–8 Simplify. The solution is (–8, 5).

  4. The Substitution Method ? ? 3 (–8) + 4 (5)  –4 –8 + 2 (5)  2 CHECK 3x + 4y  – 4Equation 1 x + 2y  2Equation 2 Solve the linear system using the substitution method. Check the solution by substituting back into the original equation. 3x + 4y  – 4 x + 2y  2 Write original equations. Substitute x and y. Solution checks. –4  –4 2  2

  5. USING ALGEBRAIC METHODS TO SOLVE SYSTEMS CHOOSING A METHOD In the first step of the previous example, you could have solved for either x or y in either Equation 1 or Equation 2. It was easiest to solve for x in Equation 2 because the x-coefficient was 1. In general you should solve for a variable whose coefficient is 1 or –1. If neither variable has a coefficient of 1 or –1, you can still use substitution. In such cases, however, the linear combination method may be better. The goal of this method is to add the equations to obtain an equation in one variable.

  6. USING ALGEBRAIC METHODS TO SOLVE SYSTEMS THE LINEAR COMBINATION METHOD 1 2 3 Multiply one or both equations by a constant to obtain coefficients that differ only in sign for one of the variables. Add revised equations from Step 1. Combine like terms to eliminate one of the variables. Solve for remaining variable. Substitute value obtained in Step 2 into either original equation and solve for other variable.

  7. The Linear Combination Method: Multiplying One Equation 2x – 4y  13Equation 1 4x – 5y  8Equation 2 Solve the linear system using thelinear combination method. SOLUTION Multiply the first equation by–2so that x-coefficients differ only in sign. • –2 2x –4y  13 –4x+8y  – 26 4x –5y  8 4x– 5y  8 3y  –18 Add the revised equations and solve fory. y–6

  8. The Linear Combination Method: Multiplying One Equation Add the revised equations and solve fory. y–6 You can check the solution algebraically using the method shown in the previous example. CHECK 11 x – 2 ( ) 11 The solution is –, –6 . 2 2x – 4y  13Equation 1 4x – 5y  8Equation 2 Solve the linear system using thelinear combination method. Substitute the value ofyinto one of the original equations. 2x – 4y  13 Write Equation 1. Substitute –6for y. 2x – 4(–6)  13 2x + 24  13 Simplify. Solve for x.

  9. The Linear Combination Method: Multiplying Both Equations 7x – 12y  –22Equation 1 – 5x + 8y  14Equation 2 Solve the linear system using thelinear combination method. SOLUTION Multiplythe first equation by2and the second equation by3so that the coefficients ofydiffer only in sign. • 2 7x – 12y  –22 14x–24y  –44 –15x+ 24y  42 • 3 –5x + 8y  14 –x–2 Addthe revised equations and solve for x. x 2

  10. The Linear Combination Method: Multiplying Both Equations x 2 Add the revised equations and solve for x. 7x – 12y  –22Equation 1 – 5x + 8y  14Equation 2 Solve the linear system using thelinear combination method. Substitute the value of xinto one of the original equations. Solve for y. –5x + 8y  14 Write Equation 2. Substitute 2for x. –5 (2) + 8y  14 y= 3 Solve for y. The solution is (2, 3). Check the solution algebraically or graphically.

  11. Linear Systems with Many or No Solutions x – 2y  3 2x – 4y  7 Solve the linear system SOLUTION Since the coefficient ofxin the first equation is 1, use substitution. Solve the first equation for x. x – 2y  3 x 2y+ 3

  12. Linear Systems with Many or No Solutions Solve the first equation for x. x 2y+ 3 x – 2y  3 2x – 4y  7 Solve the linear system Substitute the expression forxinto the second equation. 2x – 4y  7 Write second equation. 2(2y+ 3) – 4y  7 Substitute 2y+ 3forx. 6  7 Simplify. Because the statement 6 = 7 is never true, there is no solution.

  13. Linear Systems with Many or No Solutions 6x – 10y  12 –15x + 25y  –30 Solve the linear system SOLUTION Since no coefficient is 1 or –1, use the linear combination method. 30x– 50y 60 •5 6x – 10y  12 –30x+ 50y –60 •2 –15x + 25y  –30 0  0 Add the revised equations. Because the equation 0 = 0 is always true, there are infinitely many solutions.

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