# Linear Equations

A linear equation or first-degree in $x$ is written in standard form as
$ax + b = 0$   with $a \ne 0$

This is solved as follows:
$ax + b = 0$
$ax = – b$   (we subtracted $b$ from both sides)
$x = – \frac{b}{a}$     (we divided both sides by $a$)

In many cases, simple first-degree equations can be solved mentally.

For example, the solution of $5x = 10$  is $x = 2$ and the solution of $2x + 3 = 0$ is $x = – \frac{3}{2}$

Example:
Solve the linear equation $29 – 2x = 15x – 5$

Solution:
We have
$29 – 2x = 15x – 5$
$29 – 2x – 15x = – 5$ (we subtracted $15x$ from both sides)
$29 – 17x = – 5$ (we combined like terms)
$– 17x = – 34$ (we subtracted $29$ from both sides)
$17x = 34$ (we multiplied both sides by $– 1$)
$x = \frac{{34}}{{17}}$ (we divided both sides by $17$)
$x = 2$

To guard against errors in arithmetics or algebra, it’s a good idea to check the solution by substituting it back into the original equation. Thus, if we substitute $x = 2$ in the equation $29 – 2x = 15x – 5$, we obtain $29 – 2\left( 2 \right) = 15\left( 2 \right) – 5$, $25 = 25$. This shows that $2$ is indeed the solution.

Example:
Solve the linear equation $\frac{1}{{y\left( {y – 1} \right)}} – \frac{1}{y} = \frac{1}{{y – 1}}$

Solution:

Multiplying both sides of the equation by the LCD $y\left( {y – 1} \right)$ and simplifying, we have
$y\left( {y – 1} \right)\frac{1}{{y\left( {y – 1} \right)}} – y\left( {y – 1} \right)\frac{1}{y} = y\left( {y – 1} \right)\frac{1}{{y – 1}}$

That is, $1 – \left( {y – 1} \right) = y$   or   $2 – y = y$

Adding $y$ to both sides of the last equation, we obtain
$2 = 2y$;    that is    $2y = 2$

From which it follows that $y = 1$. We now check by substituting $y = 1$ into the original equation to obtain
$\frac{1}{{1\left( {1 – 1} \right)}} – \frac{1}{1} = \frac{1}{{1 – 1}}$

an equation in which neither side is defined because of the zeros in the denominators. In other words, the substitution $y = 1$ doesn’t make the equation true, it makes the equation meaningless. Here, the correct conclusion is that the original equation has no root.