There are three equations on the given.
Tristan ==> x + 15y = 40
Gwen ==> x + 10y = 30
Keith ==> x + 10y = 30
Gwen and Keith's are practically the same equation which basically leaves us
x + 15y = 40 and x + 10y = 30
Rearrange this equation so that they are in the slope intercept form, y = mx + b.
[tex]\begin{gathered} x+15y=40 \\ 15y=-x+40 \\ \frac{\cancel{15}y}{\cancel{15}}=\frac{-x+40}{15} \\ y=-\frac{x}{15}+\frac{8}{3} \\ \\ x+10y=30 \\ 10y=-x+30 \\ \frac{\cancel{10}y}{\cancel{10}}=-\frac{x}{10}+\frac{30}{10} \\ y=-\frac{x}{10}+3 \end{gathered}[/tex]Now that we have two equation, use any values of x for the two equations, and plot the value in the graph.
Plotting for y = -x/15+8/3
[tex]\begin{gathered} \text{Use }x=10,\text{ and }x=25 \\ \text{if }x=10 \\ y=-\frac{10}{15}+\frac{8}{3} \\ y=2\rightarrow\text{plot coordinates }(10,2) \\ \\ \text{if }x=30 \\ y=-\frac{25}{15}+\frac{8}{3} \\ y=1,\text{ plot coordinate }(25,1) \end{gathered}[/tex]Now that we have two points, plot the two points in the graph, and connect a line to form the line for the equation x + 15y = 40 (original equation)
Plotting for y = -x/10 + 3
[tex]\begin{gathered} \text{Use }x=0\text{ and }x=30 \\ \\ \text{if }x=0 \\ y=-\frac{0}{10}+3 \\ y=3,\text{ plot coordinate }(0,3) \\ \\ \text{if }x=30 \\ y=-\frac{30}{10}+3 \\ y=-3+3 \\ y=0,\text{ plot coordinate }(30,0) \end{gathered}[/tex]Do the same as the previous equation and stack it on top of the previous line.
The intersection of the two lines is the solution. The intersection is at (10,2) which means
x = 10, and y = 2.
