It’s easier to put in \(j\) and \(d\) so we can remember what they stand for when we get the answers.

There are several ways to solve systems; we’ll talk about graphing first.

When equations have no solutions, they are called inconsistent equations, since we can never get a solution.

Here are graphs of inconsistent and dependent equations that were created on the graphing calculator: Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns with two equations.

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).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).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.We can see the two graphs intercept at the point \((4,2)\). Push ENTER one more time, and you will get the point of intersection on the bottom! Substitution is the favorite way to solve for many students!This means that the numbers that work for both equations is We can see the two graphs intercept at the point \((4,2)\). It involves exactly what it says: substituting one variable in another equation so that you only have one variable in that equation.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.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.

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