Given the system of equations;
[tex]\begin{gathered} x+3y=6---(1) \\ y=\frac{1}{2}x+7---(2) \end{gathered}[/tex]We shall first of all re-arrange the equations in the slope-intercept form;
[tex]y=mx+b[/tex]Note that the second one has already been expressed in the slope-intercept form. For the first one we would now have;
[tex]\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}[/tex]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;
[tex]\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}[/tex]Hence, for the first equation we have the two points;
[tex]\begin{gathered} A(0,2) \\ B(6,0) \end{gathered}[/tex]For the second equation;
[tex]\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}[/tex]For the second equation we now have the points;
[tex]\begin{gathered} A(0,7) \\ B(-14,0) \end{gathered}[/tex]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;
[tex](-6,4)[/tex]
