Solve The Following System of Linear Equations With An Additional Parameter

By Math Vault | Algebra

Solve the System of Linear Equations with an additional parameter h

Today, Connor asked the following question involving a system of linear equation with an additional parameter:

Given that $x_1 + hx_2 = 4$  and $3x_1 + 6x_2 =8$, find all pairs of the form $(x_1,x_2)$ satisfying both equations.

OK…What do we do with that additional $h$? Well, read on!

Ignoring the h First

We are given this system of linear equations from the get-go:

\begin{cases}x_1+hx_2=4 & (a) \\ 3x_1+6x_2=8 & (b) \end{cases}

What to do? Well, we can try to solve this system of linear equations by regarding $h$ as an ordinary coefficient first, and see what it takes us to as we proceed judiciously. First, dividing the equation $(b)$ by $3$ on both sides yields the following equivalent system:

\begin{cases}x_1+hx_2=4 & (1) \\ x_1+2x_2=\frac{8}{3} & (2) \end{cases}

Turning the 2nd equation into $(2)-(1)$, we get:

\begin{align*} \begin{cases} x_1+hx_2=4 \\ (2-h)x_2=\frac{8}{3} – 4 \end{cases} \qquad & \Leftrightarrow \qquad \begin{cases} x_1+hx_2=4 & (3) \\ (h-2)x_2=4 – \frac{8}{3} = \frac{4}{3} & (4) \end{cases} \end{align*}

At this point, our latest system is still equivalent to the original one (i.e., $(a)$ and $(b) \iff (3)$ and $(4)$) . And with $(4)$, we are definitely one step closer towards solving for $x_2$. But before we do just that, notice that we can’t solve for $x_2$ by moving $h-2$ to the right, if $h-2=0$ (i.e., $h=2$). This would then open up two cans of worms…

Case 1 — When h=2

Replace $h$ with 2 in the equations $(3)$ and $(4)$, the following system pops up:

\begin{cases} x_1+2x_2=4 \\ 0= \frac{4}{3} \end{cases}

Oops. Contradiction. That can’t happen at any rate, so there is no solution in this case.

In fact, we don’t even have to get to $(3)$ and $(4)$ to see it. This is because if we just plug in 2 into $h$ by the time we got to $(1)$ and $(2)$, we would get the following system:

\begin{cases} x_1+2x_2=4 \\ x_1+2x_2=\frac{8}{3} \end{cases}

which, as we can see, is already a system whose equations contradict each other. Here — a picture is worth a thousand words:

System of Linear Equations with two parallel lines

You see…it’s because they are parallel to each other. Hence no intersection point!

Case 2 — When h ≠ 2

Once that hurdle is clear, we can move on to something more serious. For the record, here is the latest system we have so far:

\begin{cases} x_1+hx_2=4 & (3)\\ (h-2)x_2=4 – \frac{8}{3} = \frac{4}{3} & (4) \end{cases}

Since $h \ne 2$ now, $h-2 \ne 0$. We can therefore proceed to divide the equation $(4)$ by $h-2$ on both sides, in which case we get:

\begin{cases} x_1+hx_2=4 & (5) \\ x_2= \frac{4}{3(h-2)} & (6) \end{cases}

Lastly, if we turn $(5)$ into $(5)-h(6)$, we get a very neat system in return:

\begin{cases} x_1=4 – \frac{4h}{3(h-2)}=\frac{8h-24}{3h-6} & (7) \\ x_2= \frac{4}{3(h-2)}= \frac{4}{3h-6}& (8) \end{cases}

So for each $h \ne 2$, we get a unique pair of solution of the form $\left( \dfrac{8h-24}{3h-6}, \dfrac{4}{3h-6} \right)$.

Some Illustrations (for the Case where h ≠ 2)

For example, if $h=0$, then the original system becomes:

\begin{cases} x_1=4 \\ 3x_1+6x_2=8 \end{cases}

In which case, the unique solution would be:

$$\left( \frac{8h-24}{3h-6} , \frac{4}{3h-6} \right)_{h=0} = \left( 4,-\frac{2}{3} \right)$$

Here is a picture:

System of linear equations where h=0

The intersection point occurs at $(4, -\frac{2}{3})$.

And in the case where $h=10$, the original system becomes:

\begin{cases} x_1+10x_2=4 \\ 3x_1+6x_2=8 \end{cases}

In which case, the unique solution would be:

$$\left(\frac{8h-24}{3h-6}, \frac{4}{3h-6} \right)_{h=10} = \left(\frac{56}{24}, \frac{4}{24} \right) = \left( \frac{7}{3}, \frac{1}{6} \right)$$

Here is a picture again:

System of linear equations when h=10

Notice how the 1st equation (the red line) got rotated counter-clockwise about $(4,0)$ as h increases. The intersection point now occurs at $\displaystyle \left( \frac{7}{3}, \frac{1}{16} \right)$.

And this, is how we can solving infinitely many system of linear equations all at once. 😉


About the Author

Math Vault and its Redditbots has the singular goal of advocating for education in higher mathematics through digital publishing and the uncanny use of technologies. Head to the Vault for more math cookies. :)