SOLUTION: The slope of a line whose equation is (-2/3)x+5=x+y-3

Question 285860: The slope of a line whose equation is (-2/3)x+5=x+y-3
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Rearrange until you get it into slope-intercept form: y = mx+b.
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(-2/3)x+5=x+y-3
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Multiply everything by 3...
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3((-2/3)x+5)=3(x+y-3)
-2x + 15 = 3x + 3y - 9
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Subtract 3x from both sides...
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-5x + 15 = 3y -9
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Add 9 to both sides...
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-5x + 24 = 3y
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Divide both sides by 3...
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(-5/3)x + 8 = y
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y = (-5/3)x + 8
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slope = -5/3
y-intercept occurs where x=0, which is...
y = (-5/3)(0) + 8 = 8
or the point (0,8).
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You only need two points to draw a straight line, so the x-intercept is useful.
It occurs where y=0.
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0 = (-5/3)x + 8
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Multiply by 3
0 = -5x + 24
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Add 5x to both sides
5x = 24
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Divide by 5
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x = 24/5 = 4+4/5 = 4.80
which is the point (4.8,0)
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You can check the graph to see...
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