Question

Which point is the solution to the following system of equations?
x² + y² = 13
2x- y=4
(-2, -3)
(-3, -2)
(2,3)
(3, 2)​

Answer 1

The point (3, 2) is the solution to given system of equations

Solution:

Given that system of equations are:

$x^2 + y^2 = 13$    ------ eqn 1

$2x - y = 4$    ------- eqn 2

From eqn 2,

y = 2x - 4

Substitute y = 2x - 4 in eqn 1

$x^2 + (2x - 4)^2 = 13\\\\x^2 + 4x^2 + 16 - 16x = 13\\\\5x^2 -16x + 3 = 0$

Let us solve the above equation by quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the Quadratic Formula for $5x^2 -16x + 3 = 0$ where  a = 5, b = -16, and c = 3

$\begin{aligned}&x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4(5)(3)}}{2 \times 5}\\\\&x=\frac{16 \pm \sqrt{256-60}}{10}\\\\&x=\frac{16 \pm \sqrt{196}}{10}\end{aligned}$

The discriminant $b^2 - 4ac>0$ so, there are two real roots.

$\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}$

Substitute for x = 0.2 and x = 3 in 2x - y = 4

when x = 3

2(3) - y = 4

6 - y = 4

y = 2

when x = 0.2

2(0.2) - y = 4

0.4 - y = 4

y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)