# Graphing systems of linear equations.

I'm currently finishing the unit systems of linear equations and I ran into trouble while attempting to read the the table of values. I am able to graph systems of equations and find solutions on a graph quite easily but for some reason I get lost when it comes to tables, I think its because I've never really done it before. With the following table of values I have to state whether or not it includes a solution to the system of linear equations it represents.

I'm confused as to how each column would look in slope intercept form. However, when there is only a x and y column I'm assuming you can just plot the points and find the slope to then determine if there is a solution to the system. Here is an example of what I'm talking about:

I think that what you're supposed to recognize if that as $x$ increases by $2$, $y_A$ increases by $3$ and $y_B$ increases by $4$.

This suggests that $$m_A=\frac{\Delta y_A}{\Delta x}=\frac{3}{2}$$ and $$m_B=\frac{\Delta y_B}{\Delta x}=\frac{4}{2}=2$$

Recall that the y-intecept occurs when $x=0$, so you can read the intercepts for $y_A$ and $y_B$ right from the table (e.g. when $x=0$, $y_A=-2$ and $y_B=-5$).

The next step is to take this information and write equations (in slope intercept form). This problem gives you 3 perspectives of the solution of a system of equations (geometric, tabular, and algebraic).

The table is telling you about two linear equations \begin{align*} y_A &= m_A x + b_A \\ y_B &= m_B x + b_B \end{align*} A solution to the system would be a point $(x,y)$ satisfying both equations. In other words, $y_A = y_B$. Do you see any information in the table like that?

In your first table you have $2$ lines, given in slope intercept form by:

$$y=\dfrac{3x}{2}-2$$ $$y=2x-5$$

As the slopes are different, the lines intersect.

$$3x-4=4x-10\to x=6$$

$$y=2x+14$$ $$y=2x+4$$
The slopes are the same, and so the lines are parallel. As the $y$-intercepts are different, there are no solutions.
• yep. in your first table you can cheat and see that $x=6, Y_a=7, Y_b=7$ is there and this is your solution. – JonMark Perry Aug 18 '15 at 18:05