how do I find the two sets of points for the following algbrea equation?
y=-3x+2
The equation stands for a straight line with a slope of -3 and a y-intercept of 2.
Points on the line may be obtained by assigning a value to x, and then evaluating y according to the equation. For example, a point with x=0 gives y=-3(0)+2=2, therefore the point (0,2) is on the given line.
Similarly, for x=1, y=-3(1)+2=-1, or (1,-1) is also on the line.
Continue in the same way, assigning different values to x, you will obtain as many points as you need that lie on the given line.
To find two sets of points for the given algebraic equation, you will need to substitute different values for x and solve for y.
Let's select an arbitrary value for x and find the corresponding y. We will do this twice to get two different points.
Set 1:
Let's choose x = 0:
Substitute x = 0 into the equation: y = -3(0) + 2
Simplify the expression: y = 0 + 2 = 2
So, when x = 0, y = 2.
Therefore, the first set of points is (0, 2).
Set 2:
Now let's choose a different value for x, such as x = 1:
Substitute x = 1 into the equation: y = -3(1) + 2
Simplify the expression: y = -3 + 2 = -1
So, when x = 1, y = -1.
Therefore, the second set of points is (1, -1).
In summary, the two sets of points for the equation y = -3x + 2 are (0, 2) and (1, -1).