Hi Jennifer,
I'll look at the second problem.
I would start by plotting two special points.
First let x = 0 then since 2y + x = 3, when x = 0,
2y + 0 = 3 and hence y = ^{3}/_{2} .
Thus the point x = 0, y = ^{3}/_{2}, in other words the point (0, ^{3}/_{2} ) is on the graph.
For the second point let y = 0. Then since 2y + x = 3, when y = 0,
0 + x = 3 and hence x = 3.
Thus (3, 0) is on the graph also.
Now plot more points,
when x = 1, 2y + x = 3 so 2y + 1 = 3 and hence y = 2
when x = 1, 2y + x = 3 so 2y + 1 = 3 and hence y = 1
when x = 2, 2y + x = 3 so 2y + 2 = 3 and hence y = ^{5}/_{2}
As you plot more points you can see that they all lie on a straight line and this line is the graph of the equation 2y + x = 3.
Now try the other problem.
Penny
