document.write( "Question 268067: Two cars leave an intersection. One car travels east; the other north. when the car traveling east had gone 15 miles, the distance between the cars was 5 miles more than the distance traveled by the car heading north. How far had the northbound car traveled? \r
\n" ); document.write( "\n" ); document.write( "Thank you for your help. I am stumped on how to set up this problem for solving.\r
\n" ); document.write( "\n" ); document.write( "Thank you, Nora
\n" ); document.write( "

Algebra.Com's Answer #196572 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Hi Nora---
\n" ); document.write( "Lets see what we can do:
\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let d=distance that the northbound car travelled\r
\n" ); document.write( "\n" ); document.write( "Now know that we have a right triangle to deal with here and the distance between the two cars is the hypotenuse of the right triangle.
\n" ); document.write( "The base of the right triangle (car travelling east) is 15 mi; the hypotneuse is (d+5) and the other side is d(car travelling north).\r
\n" ); document.write( "\n" ); document.write( "Applying the pythagorean theorem, we have:
\n" ); document.write( "(d+5)^2=15^2+d^2 expand the left side and simplify
\n" ); document.write( "d^2+10d+25=225+d^2 subtract d^2 and also 25 from each side
\n" ); document.write( "d^2-d^2+10d+25-25=225-25+d^2-d^2 collect like terms
\n" ); document.write( "10d=200 divide each side by 10
\n" ); document.write( "d=20 mi-------------------distance travelled by northbound car\r
\n" ); document.write( "\n" ); document.write( "distance between the cars=d+5=20+5=25 mi\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "25^2=15^2+20^2
\n" ); document.write( "625=225+400
\n" ); document.write( "625=625\r
\n" ); document.write( "\n" ); document.write( "Does this help?---ptaylor\r
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