document.write( "Question 7997: I cannot understand how to even start this problem.\r
\n" ); document.write( "\n" ); document.write( "Each car in a fleet of 24 rental cars is either red or blue. There are 3 more blue cars than twice the number of red ones. How many red ones are there? \n" ); document.write( "

Algebra.Com's Answer #4390 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Well, you can always start by assigning variables to the quantities you are trying to find.\r
\n" ); document.write( "\n" ); document.write( "Let B = the number of blue cars and R = the number of red cars.\r
\n" ); document.write( "\n" ); document.write( "Now you have to translate the words of the problem into algebraic statements.\r
\n" ); document.write( "\n" ); document.write( "The fleet of red and blue cars is 24 cars. So, R + B = 24
\n" ); document.write( "There are 3 more (add 3 to)blue cars than twice (2 times) the number of red cars. Or, B + 3 = 2R\r
\n" ); document.write( "\n" ); document.write( "Now we have a system of two equations that we need to solve, Since the problem asks only for the number of red cars, we'll try and solve for R.\r
\n" ); document.write( "\n" ); document.write( "R + B = 24
\n" ); document.write( "B + 3 = 2R Rewrite this as: B = 2R - 3 and substitute into the first equation.\r
\n" ); document.write( "\n" ); document.write( "R + (2R - 3) = 24 Simplify and solve for R.
\n" ); document.write( "3R - 3 = 24 Add 3 to both sides.
\n" ); document.write( "3R = 27 Divide both sides by 3.
\n" ); document.write( "R = 9 There are 9 red cars.\r
\n" ); document.write( "\n" ); document.write( "If you wanted to find out how many blue cars there are, take 24 - 9 = 15 blue cars. \n" ); document.write( "

\n" );