Simultaneous Quadratic Equations: $x^2 + y ^ 2 - 2 x + 6y - 35 = 0$ and $2x + 3y = 5$

Question

I've been given the task to simultaneously solve:

$$x^2 + y ^ 2 - 2 x + 6y - 35 = 0$$

$$2x + 3y = 5$$

I've tried applying the substitution method by reordering the second equation to both $x$ and $y$, but both times I have got a surd.

According to the answers I should be getting $(-2, 3)$ and $(\frac{100}{13}, -\frac{45}{13})$, so evidently I am doing something wrong.

This is in the chapter Equations Reducible to Quadratics and I can not see how the material in the intro applies to the question.

Answer 1

From the second equation, we have that $x=\frac{5-3y}{2}$. Plugging this into the first equation, $$\left(\frac{5-3y}{2}\right)^2+y^2-(5-3y)+6y-35=0\\ \frac{9y^2-30y+25}{4}+y^2+9y-40=0$$ Multiplying by $4$, $$9y^2-30y+25+4y^2+36y-160=0\\ 13y^2+6y-135=0$$ Applying the quadratic formula, $$y=\frac{-6\pm\sqrt{6^2-4(13)(-135)}}{26}\\ =\frac{-6\pm84}{26}=\frac{-3\pm42}{13}=3,-\frac{45}{13}$$ Plugging these values back into our equation $x=\frac{5-3y}{2}$, we get solutions of $(2,-3)$, and $(\frac{100}{13},-\frac{45}{13}).$

Answer 2

i will write the first equation as $$4x^2 + 4y^2 -8x + 24y -140 = 0 \tag 1$$ form the second equation, we have $$2x = 5 - 3y, 4x^2 = (5-3y)^2 = 9y^2 - 30y + 25 \tag 2$$ subbing $(2)$ in $(1)$ gives us $$(9y^2-30y+25)+ 4y^2-4(5-3y)+24y - 140 = 0$$ this simplifies to $$0=13y^2+6y-135= (y-3)(13y+45) $$

i hope you can take it from here.

Answer 3

$x^2 + y^2 + 3y - 5 + 6y - 35 = 0 \to x^2 = 40 - y^2 - 9y \to (5-3y)^2 = 4x^2 = 160 - 4y^2 - 36y \to 25 - 30y + 9y^2 = 160 - 4y^2 - 36y \to 13y^2 + 6y - 135 = 0$. Can you continue?

Answer 4

By completing the square you can rewrite the first equation as $$(x-1)^2+(y+3)^2=45.$$ So after the substitution $s=x-1$, $t=y+3$ you have new equations (which may be a bit simpler) $$ \begin{align*} s^2+t^2&=45\\ 2s+3t&=12 \end{align*} $$

This system of equations has a geometrical interpretation as intersection of a circle and a line. So you can draw a picture which might help you visualize the problem.

Now if you express $t=4-\frac23s$ from the second equation you get $$t^2=16-\frac{16}3s+\frac49s^2.$$ Substituting this into the first equation you get $$\frac{13}9s^2-\frac{16}3s-29=0\\ 13s^2-48s-261=0$$

This quadratic equation has two solutions $s=-3$ and $s=\frac{87}{13}$, which lead to $x=-2$ and $x=\frac{110}{13}$.
If you know values of $s$ and $x$, you can easily calculate $t$ and $y$.