Solving Systems Of Equations Problems

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This will help us decide what variables (unknowns) to use.

What we want to know is how many pairs of jeans we want to buy (let’s say “\(j\)”) and how many dresses we want to buy (let’s say “\(d\)”).

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.

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).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!When there is only one solution, the system is called independent, since they cross at only one point.When equations have infinite solutions, they are the same equation, are consistent, and are called dependent or coincident (think of one just sitting on top of the other).If you're seeing this message, it means we're having trouble loading external resources on our website.If you're behind a web filter, please make sure that the domains *.and *.are unblocked.Push \(Y=\) and enter the two equations in \(=\) and \(=\), respectively. When you get the answer for \(j\), plug this back in the easier equation to get \(d\): \(\displaystyle d=-(4) 6=2\). You’ll want to pick the variable that’s most easily solved for.Note that we don’t have to simplify the equations before we have to put them in the calculator. You may need to hit “ZOOM 6” (Zoom Standard) and/or “ZOOM 0” (Zoom Fit) to make sure you see the lines crossing in the graph. Let’s try another substitution problem that’s a little bit different: Probably the most useful way to solve systems is using linear combination, or linear elimination.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.Remember these are because of the Additive Property of Equality, Subtraction Property of Equality, Multiplicative Property of Equality, and Division Property of Equality: \(\displaystyle \begin\color\\\,\left( \right)\left( \right)=\left( \right)6\text\\,\,\,\,-25j-25d\,=-150\,\\,\,\,\,\,\underline\text\\,\,\,0j 25d=\,50\\25d\,=\,50\d=2\\d j\,\,=\,\,6\\,2 j=6\j=4\end\). \(\displaystyle \begin\color\,\,\,\,\,\,\,\text-3\\color\text\,\,\,\,\,\,\,\text5\end\) \(\displaystyle \begin-6x-15y=3\,\\,\underline\text\\,29x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=58\\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\2(2) 5y=-1\\,\,\,\,\,\,4 5y=-1\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5y=-5\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-1\end\) ). In the example above, we found one unique solution to the set of equations.

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