# Solving Problems With Linear Systems

We can do this for the first equation too, or just solve for “\(d\)”.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!

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Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together.

Let's look at an example: Find the ordered pair for which Thus, the solution to the system is .

In this case both equations have "y" so let's try subtracting the whole second equation from the first: So now we know the lines cross at x=1.

Note that we solve Algebra Word Problems without Systems here, and we solve systems using matrices in the Matrices and Solving Systems with Matrices section here.

A system of linear equations is where all of the variables are to the power 1.

There are three elementary ways to solve a system of linear equations.

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.

The easiest way for the second equation would be the intercept method; when we put for the “\(d\)” intercept.

“Systems of equations” just means that we are dealing with more than one equation and variable.

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

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