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.