Money and Mixture Problems with two variables.

[bahaghari07]

I'm having big big trouble understanding this problem, and i know it's cake for some or even most you... but mixture problems are one of my weaknesses. (i have alot, you see...)

Please, please, please! Help me!

The total value of the dimes and quarters in a coin bank is $3.70. If the quarters were dimes and the dimes were quarters. the total value of the coins would be $4. Find the number of dimes and the number of quarters in the bank

So far, .10d + .25q=3.70 and .25d + .10q = 4 is all i've got to show for this problem.

 

[Guest]

Simplify one of the equations so that you have one side with one and only one variable(make sure to also deal with the number in front). Do the same to the other equation, leaving the same variable on its own. You've set it up well; the d in one equation will have the same value as in the other and the same goes for the q. Now look at the two equations, and what do you see that will allow you to only deal with a single equation?

If you can get to that point, you should be able to do it.

 

[soroban]

Hello, bahaghari07!

The total value of the dimes and quarters in a coin bank is $3.70.
If the quarters were dimes and the dimes were quarters. the total value of the coins would be $4.
Find the number of dimes and the number of quarters in the bank.

So far: \(\displaystyle \:0.10d\,+\,0.25q\:=\:3.70\,\) and \(\displaystyle \,0.25d\,+\,0.10q\:=\:4\;\) . . . Good!

You have a system of equations . . . solve it.

You may want to eliminate those awful decimals; multiply the equations by 20:

. . \(\displaystyle \begin{array}{ccc}0.10d\,+\,0.25q\:=\:3.70 & \;\Rightarrow\; & 2d\,+\,5q\:=\:74 \\ 0.25d\,+\,0.10q\:=\:4.00 & \;\Rightarrow\; & 5d\,+\,2q\:=\:80\end{array}\)

Go for it!