Question

Solve this system of equations by graphing. First graph the equations, and then type the solution.x+3y=6y=1/2x+7

Answer 1

Given the system of equations;

$\begin{gathered} x+3y=6---(1) \\ y=\frac{1}{2}x+7---(2) \end{gathered}$

We shall first of all re-arrange the equations in the slope-intercept form;

$y=mx+b$

Note that the second one has already been expressed in the slope-intercept form. For the first one we would now have;

$\begin{gathered} x+3y=6 \\ 3y=-x+6 \\ \text{Divide both sides by 3;} \\ \frac{3y}{3}=-\frac{x}{3}+\frac{6}{3} \\ y=-\frac{1}{3}x+2 \end{gathered}$

To plot this equations on a graph we take two extreme points. We can do this by finding the value of y when x = 0, and y when x = 0.

For the first equation, we would have;

$\begin{gathered} y=-\frac{1}{3}x+2 \\ \text{When x}=0 \\ y=-\frac{1}{3}(0)+2 \\ y=0+2 \\ y=2 \\ \text{That means we have the point }(0,2) \\ \text{Also, when y}=0 \\ 0=-\frac{1}{3}x+2 \\ \frac{1}{3}x=2 \\ \text{Cross multiply this and you'll have;} \\ x=2\times3 \\ x=6 \\ We\text{ now have our second point, }(6,0) \end{gathered}$

Hence, for the first equation we have the two points;

$\begin{gathered} A(0,2) \\ B(6,0) \end{gathered}$

For the second equation;

$\begin{gathered} y=\frac{1}{2}x+7 \\ \text{When x}=0 \\ y=\frac{1}{2}(0)+7 \\ y=0+7 \\ y=7 \\ \text{This means we have the point }(0,7) \\ \text{Also, when y}=0 \\ 0=\frac{1}{2}x+7 \\ -\frac{1}{2}x=7 \\ \text{Cross multiply and you'll have;} \\ x=7\times(-2) \\ x=-14 \\ \text{That means we now have the second point which is }(-14,0) \end{gathered}$

For the second equation we now have the points;

$\begin{gathered} A(0,7) \\ B(-14,0) \end{gathered}$

We can now input both sets of coordinates and our graph would come out as shown below.

The point of intersection as we can see is at where x = -6 and y = 4. Therefore;

ANSWER:

The solution to the system of equations as shown on the graph is;

$(-6,4)$