SOLUTION: 5x-40=6y 2y=8-3x by the addition method

Question 284175: 5x-40=6y
2y=8-3x by the addition method
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
The 'addition' method for solving systems of linear equations is also known as the 'elimination' method.
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First, get the equations into standard form: Ax + By = C
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5x -40 = 6y
5x - 6y = 40 :: (EQ 1)
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2y = 8 - 3x
3x + 2y = 8 :: (EQ 2)
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If we multiply EQ 2 by 3, the two equations will be in shape to eliminate the 'y' term.
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5x -6y = 40 :: (EQ 1)
9x + 6y = 24 :: (EQ 2 * 3)
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adding...
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14x = 64
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x = 64/14 = 32/7
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substitute x = 32/7 in EQ 1 to find y
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5(32/7) -6y = 40
160/7 -6y = 40
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multiply both sides by 7 to remove the denominator
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160 -42y = 280
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subtract 120 from both sides
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-42y = 120
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divide both sides by -42
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y = -120/42 = -60/21
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So, we think the intersection of the two lines will be: (32/7, -60/21) = (4 4/7, -2 6/7)
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Graphing is good way to check.
The slope-intercept form is a good way to compute the lines: y = mx+b.
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5x-40=6y
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6y = 5x -40
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divide by 6
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y = 5/6x - 40/6
y = 5/6x - 20/3
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2y=8-3x
2y = -3x + 8
y = -3/2x + 4
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