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
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