document.write( "Question 81665: Graphs/81664 (2007-05-11 11:05:43): I can't seem to fully understand how to get through the whole process to the answer.\r
\n" ); document.write( "\n" ); document.write( "in the book the problem states as follows:
\n" ); document.write( "5y-3x=15
\n" ); document.write( "let y=0 5y-3x =15
\n" ); document.write( "5(0)-3x =15
\n" ); document.write( "-3x =15
\n" ); document.write( "x =15
\n" ); document.write( "let x=0 5y-3x =15
\n" ); document.write( "5y-3(0)=15
\n" ); document.write( "5y =15
\n" ); document.write( "y =3
\n" ); document.write( "let y=6 5(6)-3x =15
\n" ); document.write( "30-3x =15
\n" ); document.write( "-3x = - 15
\n" ); document.write( "- 15
\n" ); document.write( "x = -15/-3 or 5
\n" ); document.write( "okay, this is the example in the book. I don’t understand where the uses of the 6 as y comes in. Please help. thank you.
\n" ); document.write( "0 solutions \n" ); document.write( "

Algebra.Com's Answer #58497 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
When you plug in any y value and then solve for x, you are a point that is on the line $\"5y-3x=15\"$ . So when you plug in y=6 and find x=5, that means the point (5,6) is on the line $\"5y-3x=15\"$ . What's probably throwing you off is that you're doing this backwards where you plugged in x to find y before. \n" ); document.write( "

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