Equations with the unknown in the denominator

Algebra, grade 9 Lesson 15 of 26Learning

Equations with the unknown in the denominator

In an equation like $\displaystyle {\frac{12}{x}}={4}$ the unknown is in a denominator. We solve it by getting rid of the denominators: multiplying both sides by them. Before that we find the domain, because a number that makes a denominator 0 cannot be a solution.

Solving with the unknown in the denominator: the steps

Factor the denominators.
Find the domain: the numbers that make a denominator 0 are left out.
Multiply both sides by the common denominator – the fractions disappear.
Solve the equation that is left.
Check each solution against the domain: a solution outside it is rejected.

An equation with one fraction

$\displaystyle {\frac{x+2}{x-1}}={4}$ .
The domain: $\displaystyle x\neq1$ . Multiply both sides by $\displaystyle x-1$ :

\begin{aligned}\frac{x+2}{x-1}&=4 &&/\cdot(x-1)\\ x+2&=4x-4 &&/-x+4\\ 6&=3x &&/:3\\ x&=2\end{aligned}

$\displaystyle 2$ is in the domain, so the solution is $\displaystyle {x}={2}$ . Check: $\displaystyle {\frac{2+2}{2-1}}={4}$ – true.

A rejected solution

Multiplying by

\displaystyle x(x-2)

– which is 0 at

\displaystyle {x}={0}

and at

\displaystyle {x}={2}

– can add a solution the original equation does not have. So check every solution against the domain.

$\displaystyle {\frac{x+1}{x-2}}={\frac{6}{x^{2}-2x}}$ .
Factor the denominator: $\displaystyle {x^{2}-2x}={x(x-2)}$ . The domain: $\displaystyle x\neq0$ , $\displaystyle x\neq2$ . The common denominator is $\displaystyle x(x-2)$ :

\begin{aligned}\frac{x+1}{x-2}&=\frac{6}{x(x-2)} &&/\cdot x(x-2)\\ x(x+1)&=6\\ x^{2}+x-6&=0\\ (x+3)(x-2)&=0\end{aligned}

$\displaystyle {x}={-3}$ or $\displaystyle {x}={2}$ . But $\displaystyle 2$ is outside the domain – for $\displaystyle {x}={2}$ both denominators are 0 – so it is rejected. The equation has one solution: $\displaystyle {x}={-3}$ .

Exercise 1

Solve.
Write the domain, and check each solution against it.

Exercise 2

Here the equation left after multiplying is quadratic: solve it by factoring.

Formulas: algebra, grade 9

Open as a page

The laws of powers

Multiplying powers with the same base
$\displaystyle a^{m}\cdot a^{n}$ $\displaystyle =$ $\displaystyle a^{m+n}$
Dividing powers with the same base
$\displaystyle a^{m}:a^{n}$ $\displaystyle =$ $\displaystyle a^{m-n}$ $\displaystyle a\neq0$
A power of a power
$\displaystyle \left(a^{m}\right)^{n}$ $\displaystyle =$ $\displaystyle a^{m\cdot n}$
A power of a product
$\displaystyle (a\cdot b)^{n}$ $\displaystyle =$ $\displaystyle a^{n}\cdot b^{n}$
A power of a quotient
$\displaystyle \left(\frac{a}{b}\right)^{n}$ $\displaystyle =$ $\displaystyle \frac{a^{n}}{b^{n}}$ $\displaystyle b\neq0$
The power zero
$\displaystyle a^{0}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle a\neq0$

The laws of square roots

The root of a product
$\displaystyle \sqrt{a\cdot b}$ $\displaystyle =$ $\displaystyle \sqrt{a}\cdot\sqrt{b}$ $\displaystyle a\ge0,\ b\ge0$
The root of a quotient
$\displaystyle \sqrt{\frac{a}{b}}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{a}}{\sqrt{b}}$ $\displaystyle a\ge0,\ b>0$
The square of a root
$\displaystyle \left(\sqrt{a}\right)^{2}$ $\displaystyle =$ $\displaystyle a$ $\displaystyle a\ge0$
The root of a square
$\displaystyle \sqrt{a^{2}}$ $\displaystyle =$ $\displaystyle a$ $\displaystyle a\ge0$

The extended distributive law and the short multiplication formulas

The extended distributive law
$\displaystyle (a+b)(c+d)$ $\displaystyle =$ $\displaystyle ac+ad+bc+bd$
The square of a sum
$\displaystyle (a+b)^{2}$ $\displaystyle =$ $\displaystyle a^{2}+2ab+b^{2}$
The square of a difference
$\displaystyle (a-b)^{2}$ $\displaystyle =$ $\displaystyle a^{2}-2ab+b^{2}$
The difference of squares
$\displaystyle (a+b)(a-b)$ $\displaystyle =$ $\displaystyle a^{2}-b^{2}$

The parabola

Standard form
$\displaystyle y=ax^{2}+bx+c$
Vertex form – the vertex is $\displaystyle (b,k)$
$\displaystyle y=a(x-b)^{2}+k$
Product form – it meets the $\displaystyle x$ -axis at $\displaystyle x_1$ and $\displaystyle x_2$
$\displaystyle y=a(x-x_1)(x-x_2)$
The vertex’s $\displaystyle x$ , in standard form
$\displaystyle x=-\frac{b}{2a}$

The quadratic equation

The roots of $\displaystyle {ax^{2}+bx+c}={0}$ , when $\displaystyle a\neq0$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$