19.04 Non-linear Simultaneous Equations (Extended)
We solve non-linear simultaneous equations the same way as we solve linear simultaneous equations. The most significant difference is that non-linear simultaneous equations can have more than one solution.
Worked example
Solve the following system of simultaneous equations for $x$x and $y$y:
$x+y=12$x+y=12
$x^2+y=24$x2+y=24
Think: We have the equations of a line and a parabola. We can solve the equations simultaneous by using the substitution method.
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:
$y=12-x$y=12−x
Substituting this into the second equation gives:
$x^2+12-x=24$x2+12−x=24
Now we have one equation in one variable, and we can solve it for $x$x:
| $x^2+12-x$x2+12−x | $=$= | $24$24 |
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| $x^2-x-12$x2−x−12 | $=$= | $0$0 |
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Now we have a quadratic equation in $x$x. We want numbers which add to $-1$−1 and multiply to $-12$−12. These numbers are $-4$−4 and $3$3, so we can factorise the left hand side into:
| $(x-4)(x+3)$(x−4)(x+3) | $=$= | $0$0 |
The null factor law gives us two solutions:
$x=4,-3$x=4,−3
Now to find the corresponding $y$y-values, we can substitute these $x$x-values back into one of the equations. Using $y=12-x$y=12−x gives us:
$y=8,15$y=8,15
To write the whole solution set as ordered pairs we get:
$\left(x,y\right)=\left(4,8\right),\left(-3,15\right)$(x,y)=(4,8),(−3,15)
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.