When we require the solutions that are in common belonging to several equations, we say the equations are a system of equations and we speak of the simultaneous solutions to the system.
The geometric meaning of the simultaneous solutions is the set of points that belong at once to the graphs of the all of the equations in the system. If the graphs are two curves, for example, a simultaneous solution is an intersection point of the curves.
Systems of linear equations are studied in linear algebra and are solved by matrix methods or by elimination. When one or more of the equations in a system is non-linear, finding solutions can be more challenging.
We consider systems of two equations, one linear and one having a hyperbola as its graph.
Example 1
The line given by $y=ax$y=ax and the hyperbola $y=\frac{1}{x}$y=1x may or may not intersect, depending on the value of the coefficient $a$a. What range of values can $a$a have to guarantee that there is a simultaneous solution to this pair of equations?
Solutions occur when $ax=\frac{1}{x}$ax=1x. We can rearrange this to $ax^2-1=0$ax2−1=0. This is a quadratic equation in $x$x. It can be solved with the help of the quadratic formula or by some other method and we find that
$x=\frac{\pm\sqrt{4a}}{2a}$x=±√4a2a
The expression $\sqrt{4a}$√4a can only be a real number if $a\ge0$a≥0 and the expression $\frac{\pm\sqrt{4a}}{2a}$±√4a2a is meaningless if $a=0$a=0. So, we must require $a>0$a>0.
Example 2
For what values of $b$b does the following system have just one solution?
| $y$y | $=$= | $-2x+b$−2x+b |
| $y$y | $=$= | $\frac{1}{x}$1x |
At an intersection point, $-2x+b=\frac{1}{x}$−2x+b=1x. This equation can be rearranged to $2x^2-bx+1=0$2x2−bx+1=0. The solutions to this quadratic equation are given by $x=\frac{b\pm\sqrt{b^2-8}}{4}$x=b±√b2−84.
The solutions will be real numbers when $b^2-8\ge0$b2−8≥0 and there will be exactly one solution for $x$x when $b^2-8=0$b2−8=0. Thus, we require $b=\pm2\sqrt{2}$b=±2√2.
The situation is shown graphically below.
Example 3
Solve the following system:
| $y$y | $=$= | $\frac{2}{x-1}-5$2x−1−5 |
| $y$y | $=$= | $-x-1$−x−1 |
The first equation represents a hyperbola, expanded, shifted to the right and shifted vertically. However, the steps in solving the system are the same as before.
We put $\frac{2}{x-1}-5=-x-1$2x−1−5=−x−1.
Then,
| $2-5(x-1)$2−5(x−1) | $=$= | $(-x-1)(x-1)$(−x−1)(x−1) |
| $7-5x$7−5x | $=$= | $-x^2+1$−x2+1 |
| $x^2-5x+6$x2−5x+6 | $=$= | $0$0 |
| $(x-3)(x-2)$(x−3)(x−2) | $=$= | $0$0 |
| $x$x | $=$= | $3\ \ \text{or}\ \ x\ =\ 2$3 or x = 2 |
Substituting the value $x=3$x=3 into either of the two equations in the system gives $y=-4$y=−4, and substituting $x=2$x=2 gives $y=-3$y=−3. Therefore, the solutions are the points $(3,-4)$(3,−4) and $(2,-3)$(2,−3).
Worked examples
Question 1
Consider the hyperbola $xy=2$xy=2 and the straight line $y=x+6$y=x+6.
The graph of the hyperbola $xy=2$xy=2 is given. On the same set of axes, graph $y=x+6$y=x+6.
Loading Graph...How many points of intersection are there between the hyperbola $xy=2$xy=2 and the line $y=x+6$y=x+6.
Question 2
Consider the following system of equations:
| Equation 1 | $x-y=13$x−y=13 |
| Equation 2 | $xy=48$xy=48 |
First solve for $x$x.
For $x=16$x=16, find $y$y.
For $x=-3$x=−3, find $y$y.
Question 3
Consider the hyperbola $y=\frac{3}{x}$y=3x and the line $y=5$y=5.
To solve for the point of intersection of the hyperbola and the line, Laura forms the equation $\frac{3}{x}=5$3x=5. At how many points will the two graphs intersect?
Solve for the $x$x-coordinate of the point of intersection.
Hence state the coordinates of the point of intersection in the form $\left(x,y\right)$(x,y).