Reduction to a linear equation

Solving linear equations and inequalities: Linear equations in one unknown

Reduction to a linear equation

An equation of the form \[ax+b=0\] in which \(x\) is an unknown number, and \(a\) and \(b\) are given (known) numbers with \(a\ne0\), is called a linear equation in \(x\). A less common name is first degree equation. The equations in the previous section can all be rewritten in this form.

In some cases, you can reduce complicated equations to linear equations.

Find the exact solution of the equation \[\frac{-4\,x-2}{3-2\,x}=-4\] by reduction.

Start with the equation \[\frac{-4\,x-2}{3-2\,x}=-4\] and multiply the left- and right-hand side with \(3-2\,x\): \[-4\,x-2=-4\cdot (3-2\,x)\] Herewith it has become a first degree equation that can be solved in the previously discussed manner.
The solution is: \(x={{5}\over{6}}\)

New example

In the above example, we use the rule that the validity of an equation does not change if you multiply or divide the left- and right-hand side by the same number, provided that this number is nonzero. Because the numerical value of \(x\) while solving the equation is not yet known, you better check afterwards whether you have not obtained an incorrect solution by the chosen solution method. A final check after solving equations is a good idea anyway.

Find the exact solution of the equation \[(2 x-5)^2=4\] by reduction.

Start with the equation \[(2 x-5)^2=4\] Taking roots of the left- and right-hand side gives: \[2 x-5=\sqrt{4}\quad\mathrm{or}\quad 2 x-5=-\sqrt{4}\] Herewith it has become a set of two equations that can be solve in the previously discussed manner. The solution is (denoted with the \(\vee\) operator for "or"): \[ x={{3}\over{2}} \quad\lor\quad x={{7}\over{2}} \]

New example