# Solving Systems By Substitution Word Problems

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).Now let’s see why we can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below.

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).Now let’s see why we can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below.

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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.

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.

So, again, now we have three equations and three unknowns (variables).

We’ll learn later how to put these in our calculator to easily solve using matrices (see the Matrices and Solving Systems with Matrices section), but for now we need to first use two of the equations to eliminate one of the variables, and then use two other equations to eliminate the same variable: Now this gets more difficult to solve, but remember that in “real life”, there are computers to do all this work!

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!

So the points of intersections satisfy both equations simultaneously.

We’ll need to put these equations into the $$y=mx b$$ ($$d=mj b$$) format, by solving for the $$d$$ (which is like the $$y$$): First of all, to graph, we had to either solve for the “$$y$$” value (“$$d$$” in our case) like we did above, or use the cover-up, or intercept method.

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