Given the linear equation y=2/3x+1, find the y-coordinates of the points (-6, ), (-3, ), and (3, ). Please show all of your work. Plot those points and graph the linear equation
someone please help me with this problem..
for each of your points, sub in x = ... into the equation.
1st point (-6, ?)
so x = -6
y = (2/3)(-6) + 1
= -4 + 1
= -3
repeat those steps for the other two points.
When plotted the 3 points will lie in a straight line
To find the y-coordinates of the given points (-6, ), (-3, ), and (3, ), we can substitute the x-values into the given linear equation y = (2/3)x + 1 and solve for y.
1. For the point (-6, ):
Substitute x = -6 into the equation:
y = (2/3)(-6) + 1
y = -12/3 + 1
y = -4 + 1
y = -3
Therefore, the y-coordinate for the point (-6, ) is -3.
2. For the point (-3, ):
Substitute x = -3 into the equation:
y = (2/3)(-3) + 1
y = -6/3 + 1
y = -2 + 1
y = -1
Therefore, the y-coordinate for the point (-3, ) is -1.
3. For the point (3, ):
Substitute x = 3 into the equation:
y = (2/3)(3) + 1
y = 6/3 + 1
y = 2 + 1
y = 3
Therefore, the y-coordinate for the point (3, ) is 3.
Now, we can plot these points on a graph:
- Point (-6, -3)
- Point (-3, -1)
- Point (3, 3)
To graph the linear equation y = (2/3)x + 1, we can plot additional points by choosing different x-values and finding the corresponding y-values using the equation. Then, we can connect these points to form a straight line.
For example, let's choose x = 0:
y = (2/3)(0) + 1
y = 1
So, we have another point (0, 1).
Plot this point and connect all the points to obtain the graph of the linear equation y = (2/3)x + 1.