### One-Step Equations Containing Rational Numbers

Recall that a rational number is a real number that can be written as a quotient of any two real numbers other than zero in its divisor. This set of numbers includes the integers and non-integers. One–step equations involving rational numbers is solved in the same way as one–step equations containing integers, decimals or fractions. In fact, all these sets of numbers are subsets of rational numbers.

Let us as well recall that one–step equations require one step or operation to solve them, you can either add, subtract, multiply or divide both sides by a certain number in order to arrive at an equation in the form, $$x=k$$, where $$k$$ is any real number. Thus, the underlying principle behind the rules for solving one–step equations containing rational numbers is to separate the variable and the constant part, by placing the variable on the left side and the constant on the right side of the equation. This can be achieved by performing the opposite of whatever operation is involved on the original equation or by doing the inverse of what is being done to the variable.

Examples:

Solve for the unknown variable.

\small \begin{align*}&1.\;\; \displaystyle a+\frac{3}{5}=\frac{1}{5} \\&2.\;\;\frac{2}{3}x=\frac{1}{4}\\&3.\;\;b-1.2=\frac{1}{3}\\&4.\;\;w+\frac{1}{4}=6\\&5.\;\;3w=2\frac{4}{10} \end{align*}

Solutions:

\small \begin{align*}&1.\;\; \displaystyle a+\frac{3}{5}=\frac{1}{5} \end{align*}
To solve for $$a$$, subtract $$\displaystyle \frac{1}{5}$$ from both sides of the equation then simplify the right–hand side. Thus,

\small \begin{align*}\displaystyle a+\frac{3}{5}-\frac{1}{5}&=1\frac{2}{5}-\frac{1}{5}\\\therefore a&=1\frac{1}{5} \end{align*}

\small \begin{align*}2.\;\; \displaystyle \frac{2}{3}x=\frac{1}{4}\end{align*}
To solve for $$x$$, multiply $$\displaystyle \frac{3}{2}$$ to both sides of the equation then simplify the right–hand side. Hence,
\small \begin{align*} \displaystyle \frac{3}{2}\left ( \frac{2}{3}x \right )&=\frac{1}{4}\left ( \frac{3}{2} \right )\\ \therefore x&=\frac{3}{8}\end{align*}

\small \begin{align*} 3.\;\; \displaystyle b-1.2=\frac{1}{3}\end{align*}
To solve for $$b$$, add 1.2 to both sides of the equation then simplify the right–hand side. Thus,
\small \begin{align*} \displaystyle b&=\frac{1}{3}+1.2\\b&=\frac{1}{3}+1\frac{1}{5}&&\textup{convert 1.2 to fraction}\\b&=\frac{5}{15}+1\frac{3}{15}&&\textup{convert}\;\frac{1}{3}\; \textup{and}\;\frac{1}{5}\;\textup{into similar fractions} \\b&=1\frac{8}{15} \end{align*}

\small \begin{align*} 4.\;\; \displaystyle w+\frac{1}{4}=6 \end{align*}
To solve for $$2$$, subtract $$\displaystyle \frac{1}{4}$$ from both sides then simplify the right–hand side. Hence,
\small \begin{align*} \displaystyle w&=6-\frac{1}{4}\\\therefore w&= 5\frac{3}{4}\end{align*}

\small \begin{align*} 5.\;\; \displaystyle 3w&=2\frac{4}{10}\end{align*}
To solve for $$w$$, multiply both sides by $$\displaystyle \frac{1}{3}$$ then simplify the right–hand side. Thus,
\small \begin{align*} \displaystyle \frac{1}{3}\left (3w \right )&=\left (2\frac{4}{10} \right )\frac{1}{3}\\\therefore w&=\frac{24}{10}\frac{1}{3}=\frac{24}{30}=\frac{4}{5} \end{align*}

Practice Exercises

Solve for the unknown variable.

\small \begin{align*}&1.\;\; \displaystyle a-\frac{3}{25}=1\frac{1}{25} \\&2.\;\; b+1=-2\\&3.\;\;\frac{3}{100}a=24\\&4.\;\;\frac{1}{5}d =-3 \\&5.\;\;\frac{1}{2}h=\frac{1}{5} \end{align*}