document.write( "Question 174127This question is from textbook algebra 1
\n" ); document.write( ": Find two equations with graphs that intersect at (3,-1).Show that (3,-1)
\n" ); document.write( "makes both equations true.
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Algebra.Com's Answer #129033 by solver91311(24713)\"\" \"About 
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Pick two other points. They can be anything you like, so long as they are different from (3,-1) and you don't have all three points on the same line.\r
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\n" ); document.write( "\n" ); document.write( "(2,0) and (4,0) are convenient.\r
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\n" ); document.write( "\n" ); document.write( "Now, find the equation of the line that passes through both (3,-1) and the first point you chose using the two-point form of the line:\r
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\n" ); document.write( "\n" ); document.write( "\"y-y%5B1%5D=%28%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29%29%28x-x%5B1%5D%29\"\r
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\n" ); document.write( "\n" ); document.write( "Let \"P%5B1%5D\" = (3,-1) and \"P%5B2%5D\"= (2,0)\r
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\n" ); document.write( "\n" ); document.write( "So: \"y%2B1=%28%280%2B1%29%2F%283-2%29%29%28x-3%29\"\"y%2B1=x-3\"\"y=x-4\" is the equation for the line through the points (2,0) and (3,-1). Showing that (3,-1) makes the equation true is a simple matter of substituting 3 for x and -1 for y and showing that the result is a true statement.\r
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\n" ); document.write( "\n" ); document.write( "Now all you need to do is select another point and use the same process shown above to create the other equation and demonstrate that (3,-1) is in the solution set.
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