$x^2+y^2=1, 5x+12y+13=0$ Simultaneous Equations

Question

Can someone solve this for me and show working out? I just can't do it and I don't know why I am getting x and y wrong. It will be very much appreciated. As basic as possible as well please.

Answer 1

Note that $$ y = \frac{-1}{12}(13+5x) \qquad\text{(1)} $$ and so $$ 1 = x^2 + \frac{1}{144}(13+5x)^2 $$ Can you solve this for $x$ (using the quadratic formula, if need be) and then find $y$ using $(1)$?

Answer 2

HINT. Eliminate "x" or "y" from the linear equation and replace is in the nonlinear equation. Then, since squares of "x" and "y" appear in this last equation, shoulmd should arrive to a quadratic. I suppose you are perfectly able to continue from here.

Answer 3

Solve the second equation for one of the variables in terms of the other. Let's do $x$ in terms of $y$:

$$x = \frac{12y+13}{-5}.$$

Substitute this back into the first equation, and use the quadratic formula to solve for $y$.

Having solved for $y$, then use either equation to solve for $x$.

Answer 4

The linear equation can be rewritten:

$$\begin{pmatrix}{-5/12 \\ -13/12}\end{pmatrix} \cdot \begin{pmatrix}{x \\ y}\end{pmatrix} = 1$$

Both these vectors have length $1$, and their dot-product is $1$, so they are equal.