Quadratic Simultaneous Equations¶

Given a quadratic equation, and a linear equation, you can use the following method to solve them.

Example¶

$x^2 + y^2 = 5 \\ 3x - y = -1$

Rearrange the linear equation¶

Before you can start proceeding with solving the equations, you need to rearrange the linear equation to make $y$ the subject.

$\begin{align} 3x - y & = -1 \\ 3x & = -1 + y \\ 3x + 1 & = y \end{align}$

The subject of the linear equation is $y$ .

Substitute the linear equation¶

In order to use the quadratic formula, the equation needs to equal $0$ , and for it to only contain one variable.

$x^2 + (3x + 1)^2 = 5$

The linear equation has been substituted into the quadratic equation.

Expand the brackets¶

By this point chances are there are some brackets involved, these need to be squashed.

	$3x$	$+1$
$3x$	$9x^2$	$+3x$
$+1$	$+3x$	$+1$

$\begin{align} x^2 + 9x^2 + 6x + 1 & = 5 \\ 10x^2 + 6x + 1 & = 5 \end{align}$

The brackets in the equation were expanded, and the equation simplified.

Make the equation equal to 0¶

To be able to use the quadratic formula, the equation needs to equal $0$ . It is worth noting that the equation may well be better to solve by use of factorising or, my favourite, completing the square.

$\begin{align} 10x^2 + 6x + 1 & = 5 \\ 10x^2 + 6x - 4 & = 0 \end{align}$

The equation was made to equal $0$ .

Use the Quadratic Formula¶

I am now ready to use the Quadratic Formula to solve the equation.

$\begin{align} x & = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a} \\ & = \frac{-6 \pm\sqrt{6^2 - 4(10)(-4)}}{2(10)} \\ & = \frac{-6 \pm\sqrt{36 + 160}}{20} \\ & = \frac{-6 \pm\sqrt{196}}{20} \end{align}$

Solve¶

Now having an equation where $x$ is the subject, the equations can now be solved. The established $x$ values will need to be factored into the original linear equation, with $y$ as the subject.

$\begin{align} x & = \frac{-6 + \sqrt{196}}{20} \\ & = 0.4 \\ \\ y & = 3x + 1 \\ & = 3(0.4) + 1 \\ & = 1.2 + 1 \\ & = 2.2 \end{align}$

$\begin{align} x & = \frac{-6 - \sqrt{196}}{20} \\ & = -1 \\ \\ y & = 3x + 1 \\ & = 3(-1) + 1 \\ & = -3 + 1 \\ & = -2 \end{align}$

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