what would be the way to solve a system of equations like this one?

Question

Solve:

$xy=-30$
$x+y=13$

{15, -2} is a particular solution, but, how would I know if is the only solution, or what would be the way to solve this without "guessing" ?

Answer 1

We have in general $$(x+y)^2-4xy=(x-y)^2.$$ In our case that gives $(x-y)^2=289$, so $x-y=\pm 17$.

Now solve the system $x+y=13$, $x-y=17$ and the system $x+y=13$, $x-y=-17$.

Remark: This procedure for finding $x$ and $y$ given their sum and product goes back to Neo-Babylonian times. (Of course, algebraic notation was not used, but an equivalent algorithm was taught.)

Answer 2

Start with $$ x+y=13 \iff y=13-x $$ so that $$ -30=xy=x(13-x)=13x-x^2 \iff x^2-13x-30=0 $$ which solves to give $x=-2$ and $x=15$. The corresponding $y$ values are $15$ and $-2$.

Answer 3

The solutions of the system of equations $xy=p$ and $x+y=s$ are the roots of the quadratic equation:

$$x^2-sx+p=0$$ so in your case we solve $$x^2-13x-30=0$$ using the discriminant: $$\Delta=13^2+4\times 30=17^2$$ so the two roots are $$x_1=\frac{13+17}{2}=15\quad;\quad x_2=\frac{13-17}{2}=-2$$

Answer 4

$$xy=-30$$ $$x+y=13$$

from second equation $y=13-x$ then put on first we get $$x(13-x)=-30$$ $$x^2-13x-30=0$$ $$x_{1,2}=\frac{13\pm\sqrt{189}}{2}=\frac{13\pm17}{2}$$ $$x_1=\frac{13+17}{2}=15,x_2=\frac{13-17}{2}=-2$$ $$y_1=13-15=-2,y_2=13-(-2)=15$$ so the solutions are $(15,-2)$ and $(-2,15)$

Answer 5

The sum

$$x + y = 13 = -a$$

and the product

$$xy = -30 = b$$

gives rise to the quadratic equation

$$z^2 + az + b = 0.$$

Consequently, we have to solve the quadratic equation

$$z^2 - 13z - 30 = 0$$

for $z$, in order to get $x$ and $y$.

Since the discriminant $a^2 - 4b = {17}^2$, $z^2 - 13z - 30$ is factorable as:

$$z^2 - 13z - 30 = (z - 15)(z + 2).$$

Setting it equal to zero, we get the roots: $z = 15$ or $z = -2$.

Since the equations

$$x + y = 13$$

and

$$xy = -30$$

are symmetric in $x$ and $y$, we get the solutions:

$$x = 15, y = -2$$

or

$$x = -2, y = 15.$$

Answer 6

The other answers here show the strategies of:

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$1$) Substitution (the most general). See Kim Jong Un's answer on this question.

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$2)$ Solving a quadratic equation that has the roots $x,y$ (this uses the Vieta's formulas). See Sami Ben Romdhane's answer.

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$3)$ Using the fact that $(x+y)^2-4xy=(x-y)^2$ to have the numeric value of $x-y$. Then $(x+y)-(x-y)=2y$, an easy way to get the value of $y$. See André Nicolas' answer.

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I'm posting another method that works great for this particular problem.

Add the equations in a specific way to get:$$\begin{align}xy+2(x+y)&=-30+2\cdot 13=-4\\\iff (x+2)(y+2)&=0\end{align}$$

Thus either $x=-2$ or $y=-2$, which give the solutions $(-2,15)$ and $(15,-2)$, respectively.