Question

x^2-y=7 and y-5x=-11 The system of equations above has two solutions. The solution in Quadrant 1 is (A,B) and the solution in Quadrant 4 is (C,D) . Find the values of (A,B), (C,D)

Answer 1

By solving the system of equations, we can see that the solutions are:

(A, B) = (4, 9)

(C, D) = (1, -6).

How to solve the system of equations?

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:

$x = \frac{5 \pm \sqrt{(-5)^2 - 4*1*(4)} }{2*1}$

Solving that we get:

$x = \frac{5 \pm \sqrt{(9} }{2} \\\\x = \frac{5 \pm 3}{2}$

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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