Systems of two non-linear equations

Given a system of equations some of which are non-linear, we wish to find the points of intersection, if any exist. That is, we look for points that satisfy every equation in the system. 

This can be much more difficult than in the linear case.

In the following diagram, the graphs of three relations are shown. Two of them are non-linear. For each pair of graphs, there are two intersection points but there is only one point at which all three intersect.

In the linear case with graphs relating two variables, there could be zero, one or infinitely many points of intersection. Clearly, the situation is more complicated when we are dealing with non-linear functions and relations.

 

Example 1

Consider a parabola and a line given by the equations $y=x^2$y=x2 and $y=x-1$y=x−1 respectively. We could determine by drawing the graphs that there are no intersection points. We can also discover this algebraically.

At an intersection point, it would be true that $x^2=x-1$x2=x−1. That is, $x^2-x+1=0$x2−x+1=0. This quadratic equation has discriminant $\Delta=(-1)^2-4\times1\times1=-3$Δ=(−1)2−4×1×1=−3. Since the discriminant is negative, there are no real solutions and hence, there is no intersection.

 

Example 2

Consider two parabolas, given by $y=x^2$y=x2 and $y=\frac{x^2}{3}+1$y=x23​+1.

To find the points that simultaneously satisfy both these equations, we set $x^2=\frac{x^2}{3}+1$x2=x23​+1. This is a quadratic: $x^2-\frac{3}{2}=0$x2−32​=0 with solutions $x=\sqrt{\frac{3}{2}}$x=√32​ and $x=-\sqrt{\frac{3}{2}}$x=−√32​. 

Thus, the intersection points are $\left(\sqrt{\frac{3}{2}},\frac{3}{2}\right)$(√32​,32​) and $\left(-\sqrt{\frac{3}{2}},\frac{3}{2}\right)$(−√32​,32​). 

These intersection points could be verified using a graphics program or calculator if you wished. 

 

Example 3

Suppose we wish to find a line with gradient $1$1 that is tangent to the semicircle given by $y=\sqrt{1-x^2}$y=√1−x2.

The line has the equation $y=x+k$y=x+k and we need to find the value of $k$k such that there is exactly one point of intersection. We set $x+k=\sqrt{1-x^2}$x+k=√1−x2. 

To proceed, we must square both sides and simplify. Thus, $x^2+2kx+k^2=1-x^2$x2+2kx+k2=1−x2 and so, $2x^2+2kx+k^2-1=0$2x2+2kx+k2−1=0. 

The discriminant is $\Delta=4k^2-8\left(k^2-1\right)$Δ=4k2−8(k2−1) and to obtain exactly one root this must be equal to zero. So, $k^2-2\left(k^2-1\right)=0$k2−2(k2−1)=0.

Hence, $2-k^2=0$2−k2=0 and so, we could have $k=\sqrt{2}$k=√2 or $k=-\sqrt{2}$k=−√2. One of these solutions is spurious. It was introduced when the term $\sqrt{1-x^2}$√1−x2 was squared, since this gave the same result as would have been obtained if $-\sqrt{1-x^2}$−√1−x2 had been squared. The solution we want is the positive one.

We conclude that the tangent line to the semicircle is given by $y=x+\sqrt{2}$y=x+√2.

This can be checked by verifying that the equation $x+\sqrt{2}=\sqrt{1-x^2}$x+√2=√1−x2 does indeed have exactly one root: $x=-\frac{1}{\sqrt{2}}$x=−1√2​.

So, the point of tangency is $\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$(−1√2​,1√2​).

 

Worked examples

Question 1

Question 2

Question 3