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Published byJayson Willis Modified over 7 years ago

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SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2

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SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10 2x + y = 10 - 2x a) Y = - 2x + 10

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SUBSTITUTE Step 2: substitute -2x + 10 for y Step 3: Substitute the value of x that you equation 2, and solve for x. found into equation a). Solve for y: 3x + 4y = 12 2x + y = 10 The solution is (5.6, -1.2)

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NOW YOU TRY ONE 1) X + 3y = 5 2) -2x -4y = - 5

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SOLVING BY ELIMINATION SOLVE : 4x + 2y = 9 4x + 2y = 9 One equation has 4x - 4x + 3y = 16- 4x + 3y = 16 and the other has -4x Add to eliminate x 4x + 2y = 9 Choose one of the equations Substitute for y Solution is (, )

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-2x + 8y = -8 ONE EQUATION HAS 8y 5x – 8y = 20 AND THE OTHER HAS – 8y ADD TO ELIMINATE Y Choose one of the original equations Substitute 4 for x Solution (, ) WHAT IS THE SOLUTION FOR -2x + 8y + -8 5x – 8y = 20

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SOLVING AN EQUIVALENT SYSTEM 2x + 7y = 4 By multiplying the first 3x 3x + 5y = -5equation by 3 and the second by -2. The x terms become opposite. Add them and solve for y

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YOU TRY 3x + 7y = 15 5x + 2y = -4

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