*When there is only one solution, the system is called independent, since they cross at only one point.*

*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).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!

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.

Then we add the two equations to get “\(0j\)” and eliminate the “\(j\)” variable (thus, the name “linear elimination”). Now that we get \(d=2\), we can plug in that value in the either original equation (use the easiest! We then get the second set of equations to add, and the \(y\)’s are eliminated. Now we can plug in that value in either original equation (use the easiest! Sometimes, however, there are no solutions (when lines are parallel) or an infinite number of solutions (when the two lines are actually the same line, and one is just a “multiple” of the other) to a set of equations.

When there is at least one solution, the equations are consistent equations, since they have a solution.

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.

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.

In this type of problem, you would also have/need something like this: .

Now, since we have the same number of equations as variables, we can potentially get one solution for the system.

## Comments Solving Word Problems Using Systems Of Equations

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