To get rid of a negative exponent, move the part with the negative exponent to the other side of the_________(blank)
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By solving the system of equations, we can see that the solutions are:
(A, B) = (4, 9)
(C, D) = (1, -6).
<h3>How to solve the system of equations?</h3>Here we have the following system of equations:
x^2 - y = 7
y - 5x = -11
If we isolate the variable y in both equations, we get:
y = x^2 - 7
y = -11 + 5x
Now we can equate these two to get:
x^2 - 7 = -11 + 5x
now we have a quadratic equation, this can be rewritten as:
x^2 - 7 + 11 - 5x = 0
x^2 - 5x + 4 = 0
Using the quadratic formula, we will see that the solutions are:
Solving that we get:
So the two solutions for x are:
x = (5 + 3)/2 = 4
x = (5 - 3)/2 = 1
Evaluating the second equation in these x-values we get:
y = -11 + 5*4 = -11 + 20 = 9
y = -11 + 5*1 = -6
Then we have the coordinate pairs: (4, 9) (on the first quadrant) and (1, -6) on the fourth quadrant.
Learn more about systems of equations:
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