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  1. #1
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    Algebra story problem. Need Help ASAP

    ok...you have some $1 bills, $5 bills, and $10 bills. There are 225 bills total and they are worth $596. If you have seven times as many $1 as you do $10, how many of each bill do you have?????

    i have tried this problem over 20 times!!!!!!

    Sincerly
    armath
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  2. #2
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    Quote Originally Posted by armath View Post
    ok...you have some $1 bills, $5 bills, and $10 bills. There are 225 bills total and they are worth $596. If you have seven times as many $1 as you do $10, how many of each bill do you have?????

    i have tried this problem over 20 times!!!!!!

    Sincerly
    armath
    Let x = number of $1 bills, y = number of $5 bills and z = number of $10 bills.

    Then:

    x + y + z = 225 .... (1)

    x + 5y + 10z = 596 .... (2)

    x = 7z .... (3)

    I suggest substituting (3) into (1) and (2) to eliminate x. Now solve (1) and (2) simultaneously for z and y. Now sub the value of x into (3) to get x.
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by armath View Post
    ok...you have some $1 bills, $5 bills, and $10 bills. There are 225 bills total and they are worth $596. If you have seven times as many $1 as you do $10, how many of each bill do you have?????

    i have tried this problem over 20 times!!!!!!

    Sincerly
    armath
    You have 2 equations here, let "A" stand for the number of 1 dollar bills you have, "B" stand for the number of 5 dollar bills you have, and C stand for the number of 10 dollar bills you have.

    Then you know that the total number of bills is 225, so A + B + C = 225

    And you know that each A is worth 1 dollar, each B is worth 5 dollars, and each C is worth 10 dollars, and that you have 596 dollars total. So A + 5B + 10C = 596

    So set the equations next to eachother:
    \begin{array}{ccccc}<br />
A&+B&+C&=&225\\<br />
A&+5B&+10C&=&596\\<br />
\end{array}

    Now, we have three variables, but only 2 equations. But we know that there are 7 times as many ones as tens. This means that the number of ones is equal to 7 times the number of tens. (Think if there was 1 ten, then there would need to be 7 ones) so A = 7C. This means we can substitute out A:
    \begin{array}{ccccc}<br />
7C&+B&+C&=&225\\<br />
(7C)&+5B&+10C&=&596\\<br />
\end{array}

    Simplify
    \begin{array}{ccccc}<br />
&B&+8C&=&225\\<br />
&5B&+17C&=&596\\<br />
\end{array}

    Now you can solve this any way you like, I'll do it with substitution. Since B + 8C = 225, it we can see that B = 225-8C So lets substitute that into our other equation:
    5(225-8C) +17C = 596

    1125-40C +17C = 596

    -23C = -529

    C = 23

    Now we can plug that back into our other equations:
    \begin{array}{ccccc|cccccc}<br />
A&+B&+C&=&225& \Longrightarrow& A&+B&+23&=&225\\<br />
A&+5B&+10C&=&596&&A&+5B&+230&=&596\\<br />
\end{array}

    And simplify
    \begin{array}{cccc}<br />
A&+B&=&202\\<br />
A&+5B&=&-366\\<br />
\end{array}

    Again, you can solve this any way you like, I'll use matrices this time:
    \left(\begin{array}{cccc}<br />
A&+B&=&202\\<br />
A&+5B&=&366\\<br />
\end{array}\right)

    Subtract line 1 from line 2:
    \left(\begin{array}{cccc}<br />
A&+B&=&202\\<br />
&4B&=&164\\<br />
\end{array}\right)

    Divide line 2 by 4
    \left(\begin{array}{cccc}<br />
A&+B&=&202\\<br />
&B&=&41\\<br />
\end{array}\right)

    And substitute The value of B into the first equation:
    \left(\begin{array}{cccc}<br />
A&+41&=&202\\<br />
&B&=&41\\<br />
\end{array}\right)

    Subtract 41 from line 1
    \left(\begin{array}{cccc}<br />
A&&=&161\\<br />
&B&=&41\\<br />
\end{array}\right)

    So we have A=161, B=41, and C=23

    We can verify our answer by plugging them back into our equations:
    \begin{array}{ccccc|cccccc}<br />
A&+B&+C&=&225&\Longrightarrow&161&+41&+23&=&225\  \<br />
A&+5B&+10C&=&596&&161&+5(41)&+10(23)&=&596\\<br />
\end{array}

    You can use your calculator to verify that these answers are correct (they are, I checked them, but you should get into that practice).

    And so we only need to verify that they also meet the last criteria that A=7C, so we plug the values in: 161=7(23) and we see that this is correct as well. So we know that our answer is correct.



    Some parting advice: Make sure you set them up right, ie seven times as many ones as tens might make you want to say 7A=C, but plug in a value for A, and you will see that C is actually 7 times bigger than A, so it must be the other way around that A=7C. Second, when manipulating equations, use one to manipulate the other, you can never use an equation to manipulate itself, because you will end up with A=A or 0=0 or some other such useless equation. So if you solve for a variable in one equation, use that on the other equation.
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