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2007 - - - 9x + 223y = 2007

by Andrew
(New York)










































The graph of the equation 9x + 223y = 2007 is drawn on graph paper with each square representing one unit in each direction. How many of the 1 by 1 graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

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Feb 14, 2012
9x + 223y = 2007
by: Staff


Question:

by Andrew
(New York)


The graph of the equation 9x + 223y = 2007 is drawn on graph paper with each square representing one unit in each direction. How many of the 1 by 1 graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?



Answer:


To view a graph of the linear equation 9x + 223y = 2007, open the following link.

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http://www.solving-math-problems.com/images/linear-equation-01-2012-02-14.png


As you can see, the area below the graph which lies in the first quadrant forms a right triangle.


The answer to your question is the area of the right triangle.

A = ½ * base * height


base = x-intercept

height = y-intercept


A = ½ * (x-intercept) * (y-intercept)



The x-intercept:

9x + 223y = 2007

If y = 0

9x + 223 * 0 = 2007

9x + 0 = 2007

9x = 2007

9x/9 = 2007/9

x * (9/9) = 2007/9

x * (1) = 2007/9

x = 2007/9

x = 223

x-intercept = 223 units



The y-intercept:

9x + 223y = 2007

If x = 0

9 * 0 + 223y = 2007

0 + 223y = 2007

223y = 2007

223y/223 = 2007/223

y * (223/223) = 2007/223

y * (1) = 2007/223

y = 2007/223

y = 9

y-intercept = 9 units



A = ½ * (x-intercept) * (y-intercept)

A = ½ * (223) * (9)


A = 1003.5 units²



>>>the final answer is: 1003.5 paper squares




Thanks for writing.

Staff
www.solving-math-problems.com



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