document.write( "Question 64972: I have tried and tried to remember the formulas and have done research to refresh my memory, but it's just not working. Here's the question:\r
\n" ); document.write( "\n" ); document.write( "If you charge $1 per cup of a drink and sell 50 cups, then you raise your price to $2 per cup and sell only 25 cups, write an equation for the number of cups that you sell as a function of the price you charge. \"C\" is for # of cups and \"P\" is for the price you charge. Assume that the function is linear. \r
\n" ); document.write( "\n" ); document.write( "Thank you\r
\n" ); document.write( "\n" ); document.write( "This problem is from an assignment from a course through correspondence.
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Algebra.Com's Answer #45476 by cristiana(10)\"\" \"About 
You can put this solution on YOUR website!
All right, so the start point is that our function is linear.
\n" ); document.write( "Draw the coordinate system.
\n" ); document.write( "Let the price be x and number of cups be y.
\n" ); document.write( "We know that when x = 1, y = 50. Plot this point.
\n" ); document.write( "We also know that when x = 2, y = 25. Plot this point too.
\n" ); document.write( "Draw a line between the two plotted points.
\n" ); document.write( "Let's try to find the point-intercept equation of this line. Since all we know is the two points, we can make a system of two point-slope equations and find the slope. Let's do that:
\n" ); document.write( "point (1,50): y-50 = m(x-1); y = m(x-1) + 50
\n" ); document.write( "point (2,25): y-25 = m(x-2); y = m(x-2) + 25
\n" ); document.write( "Therefore, m = -25
\n" ); document.write( "Substitute -25 for m in any of the above two equations to get:
\n" ); document.write( "y=-25(x-1)+50
\n" ); document.write( "or
\n" ); document.write( "y=-25(x-2)+25 \r
\n" ); document.write( "\n" ); document.write( "You can now check these equations by replacing x=1 ($1/cup) and x=2 ($2/cup)...
\n" ); document.write( " \r
\n" ); document.write( "\n" ); document.write( "Hope it helped!
\n" ); document.write( "Cristiana
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