# 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**.