*It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.*The first trick in problems like this is to figure out what we want to know.

This means that the numbers that work for both equations is 4 pairs of jeans and 2 dresses! Here is the problem again: Solve for \(d\): \(\displaystyle d=-j 6\).

We can also use our graphing calculator to solve the systems of equations: Solve for \(y\,\left( d \right)\) in both equations. Plug this in for \(d\) in the second equation and solve for \(j\). Note that we could have also solved for “\(j\)” first; it really doesn’t matter.

Always write down what your variables will be: equations as shown below.

Notice that the \(j\) variable is just like the \(x\) variable and the \(d\) variable is just like the \(y\).

So far, we’ve basically just played around with the equation for a line, which is \(y=mx b\).

Now, you can always do “guess and check” to see what would work, but you might as well use algebra!

Let’s let \(j=\) the number of pair of jeans, \(d=\) the number of dresses, and \(s=\) the number of pairs of shoes we should buy.

So far we’ll have the following equations: \(\displaystyle \beginj d s=10\text\25j \text50d \,20s=260\end\) We’ll need another equation, since for three variables, we need three equations (otherwise, we’d theoretically have infinite ways to solve the problem).

Due to the nature of the mathematics on this site it is best views in landscape mode.

If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

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