$a^2x+(a-8)=(a+8)x\\\\(a+8)x=a^2x+(a-8)\ \ \ \ |subtract\ a^2x\ from\ both\ sides\\\\(a+8)x-a^2x=a-8\\\\(a+8-a^2)x=a-8\ \ \ \ |divide\ both\ sides\ by\ (a+8-a^2)\\\\\huge\boxed{x=\frac{a-8}{a+8-a^2}}\to\huge\boxed{x=\frac{8-x}{a^2-a-8}}$

6 0

3 years ago

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2a. Determine the equation of the line connecting the points (0, -1) and (2, 3) *

Sauron [17]

First create the equation y=Mx+ B

B is the y intercept so when y = 0, B = -1.

Now we have y=Mx - 1

The slope is M. We can calculate this by using the formula M = (y2 -y1) / (x2 - x1)

Use the points (0, -1) and (2,3) for these values. So y2 = 3, y1 = -1, x2 = 2, x1 = 0

Plug them into the equation and solve

M = (3 + 1) / (2 - 0)
M = 4/2 = 2

Now we have the equation y = 2x - 1

Next to figure out if the points given are on the line you take the values and plug them into your equation like so: 4 = 2(-2) - 1

2(-2) - 1 does not equal 4 so this point does not fall on this line. Follow this same procedure for the next point given.

8 0

3 years ago