2.06 Non-linear Simultaneous Equations (Extended)

Solve the following system of simultaneous equations for $x$x and $y$y:

$xy=12$xy=12

$x^2+y^2=25$x2+y2=25

Think: We have the equations of a circle and a hyperbola. We can solve the equations simultaneous by using the substitution method. Note that because of the equation of the hyperbola, we already know that $x\ne0$x≠0 and $y\ne0$y≠0.

Do: To use the substitution method, the first step is to rewrite one equation with $x$x or $y$y as the subject. Using the first equation we get:

$y=\frac{12}{x}$y=12x​

Substituting this into the second equation gives:

$x^2+\left(\frac{12}{x}\right)^2=25$x2+(12x​)2=25

Now we have one equation in one variable, and we can solve it for $x$x:

$x^2+\left(\frac{12}{x}\right)^2$x2+(12x​)2	$=$=	$25$25
$x^2+\frac{144}{x^2}$x2+144x2​	$=$=	$25$25	Expanding the square of the fraction
$x^4+144$x4+144	$=$=	$25x^2$25x2	Multiplying both sides of the equation by $x^2$x2
$x^4-25x^2+144$x4−25x2+144	$=$=	$0$0	Subtracting $16x^2$16x2 from both sides of the equation

Now we have a quartic equation in $x$x. We can treat this as a quadratic equation in $x^2$x2. In that case, we want numbers which add to $-25$−25 and multiply to $144$144. These numbers are $-9$−9 and $-16$−16, so we can factorise the left hand side into:

$\left(x^2-16\right)\left(x^2-9\right)=0$(x2−16)(x2−9)=0

Both factors are a difference of two squares, so we can factorise the expression further:

$\left(x-4\right)\left(x+4\right)\left(x-3\right)\left(x+3\right)=0$(x−4)(x+4)(x−3)(x+3)=0

The null factor law gives us four solutions:

$x=4,-4,3,-3$x=4,−4,3,−3

Now to find the corresponding $y$y-values, we can substitute these $x$x-values back into one of the equations. Using $y=\frac{12}{x}$y=12x​ gives us:

$y=3,-3,4,-4$y=3,−3,4,−4

To write the whole solution set as ordered pairs we get:

$\left(x,y\right)=\left(4,3\right),\left(-4,-3\right),\left(3,4\right),\left(-3,-4\right)$(x,y)=(4,3),(−4,−3),(3,4),(−3,−4)

Reflect: Given the nature of non-linear equations, we can't really predict how many solutions there will be, so we have to be careful to use all of the information we are given to make sure that we find all of the solutions. We can check these solutions by substituting them into the original equations.