I am going to change the problem by looking at the functions
y = 3x + 9 and y = 5x  3
The first thing I would do is to use a piece of scrap paper and draw a rough sketch using the intercepts. For the first function
y = 3x + 9
When x = 0, y = 9 so (0,9) is on the graph and when y = 0, x = 3 so (3,0) is on the graph.
Now add the second function
y = 5x  3
When x = 0, y = 3 so (0,3) is on the curve. When y = 0, x = ^{3}/_{5}
I purposefully didn't mark the xintercept of the second line on the graph and drew the line thick because I am not trying to be accurate here, I am just trying to see what ranges I should choose for the x and y values when I draw an accurate graph. What I see here is that if I draw a graph with the x values between 4 and 1, and the y values between 1 and 10 then the point where the lines cross will be on my graph.
To graph the second line I substituted x = 1 and got y = 2 to find a second point on the graph. I used this point rather than the xintercept so that I didn't have to use fractions. I also didn't mark a scale on the axes, but you will need to do this to answer part 3.
The equations are
y = 3x + 9
y = 5x  3
hence
3x + 9 = 5x  3
Simplifying gives me
8x = 12 or x = ^{2}/_{3}
Substituting this value into the first equation I get y = ^{9}/_{2} _{}
This looks approximately correct on my diagram, that is the curves intersect at x = ^{2}/_{3}, y = ^{9}/_{2}
Penny
