# One-Step Linear Equations

One-step linear equation is an equation, that requires only one action (operation) to be solved.

Basically, there are 4 kinds of such equations:

• ${x}+{3}={7}$ (solve by using subtraction)
• ${a}-{2}=-{3}$ (solve by using addition)
• ${2}{y}={1}$ (solve by using division)
• $\frac{{1}}{{4}}{x}={5}$ (solve by using multiplication)

You don't need to remember above kinds of equation. Just keep in mind, that to solve linear equation, we need to isolate variable, i.e. keep it on one side of equation and move everything else to another side.

Following examples will show how to deal with one-step linear equations.

First, let's see how to use subtraction

Example 1. Solve ${x}+{3}={10}$.

We see, that on the left side of the equation there are variable and number. We need to get rid of 3.

We can do that by subtracting 3 from both sides of equation:

${x}+{3}{\color{red}{{-{3}}}}={10}{\color{red}{{-{3}}}}$

${x}+{0}={7}$ (inverse property of addition: ${3}-{3}={3}+{\left(-{3}\right)}={0}$)

${x}={7}$ (identity property of addition: ${x}+{0}={x}$)

So, ${7}$ is root of the equation.

Example 2. Solve ${x}-\frac{{2}}{{3}}=\frac{{7}}{{3}}$.

We see, that on the left side of the equation there are variable and number. We need to get rid of $\frac{{2}}{{3}}$.

We can do that by adding $\frac{{2}}{{3}}$ to both sides of equation:

${x}-\frac{{2}}{{3}}{\color{red}{{+\frac{{2}}{{3}}}}}=\frac{{7}}{{3}}{\color{red}{{+\frac{{2}}{{3}}}}}$

${x}+{0}=\frac{{9}}{{3}}$ (inverse property of addition: $-\frac{{2}}{{3}}+\frac{{2}}{{3}}={0}$)

${x}={3}$ (identity property of addition: ${x}+{0}={x}$)

So, ${3}$ is root of the equation.

Next comes division.

Example 3. Solve ${2}{a}={8}$.

We see, that on the left side of the equation there is coeffcient ${2}$ near the variable. We need to get rid of it.

We can do that, by dividing both sides of equation by ${2}$:

$\frac{{{2}{a}}}{{{\color{red}{{{2}}}}}}=\frac{{8}}{{{\color{red}{{{2}}}}}}$

${1}\times{a}=\frac{{8}}{{{\color{red}{{{2}}}}}}$ (inverse property of multiplication: $\frac{{2}}{{2}}={2}\times\frac{{1}}{{2}}={1}$)

${a}=\frac{{8}}{{2}}$ (identity property of multiplication: ${1}\times{a}={a}$)

${a}={4}$

So, ${4}$ is root of the equation.

Example 4. Solve $\frac{{1}}{{15}}{y}=-\frac{{7}}{{30}}$.

We see, that on the left side of the equation there is coefficient near variable. We need to get rid of it.

We can do that, by multiplying both sides of equation by ${15}$:

${\color{red}{{{15}\times}}}\frac{{1}}{{15}}{y}={\color{red}{{{15}\times}}}{\left(-\frac{{7}}{{30}}\right)}$

${1}\times{y}={\color{red}{{{15}\times}}}{\left(-\frac{{7}}{{30}}\right)}$ (inverse property of multiplication: ${15}\times\frac{{1}}{{15}}={1}$ )

${y}=-\frac{{105}}{{30}}$ (identity property of multiplication: ${1}\times{y}={y}$)

${y}=-\frac{{7}}{{2}}$

So, $-\frac{{7}}{{2}}$ is root of the equation.

Note, that it doesn't matter whether we write ${15}\times\frac{{1}}{{15}}{y}$ or $\frac{{1}}{{15}}{y}\times{15}$ due to the commutative property of multiplication. Same applies to ${15}\times{\left(-\frac{{7}}{{30}}\right)}$ (we could write it as $-\frac{{7}}{{30}}\times{15}$).

Now, it is time to exercise.

Exercise 1. Solve the equation ${x}+{2}=\frac{{7}}{{2}}$.

Answer: $\frac{{3}}{{2}}={1.5}$.

Exercise 2. Find roots of the equation ${x}-{2}={5}$.

Answer: ${7}$.

Exercise 3. Solve the equation ${2}{x}={5}$.

Answer: $\frac{{5}}{{2}}={2.5}$.

Exercise 4. Solve the equation $\frac{{1}}{{3}}{a}={15}$.

Answer: ${45}$.

Exercise 5. Solve the equation $\frac{{3}}{{2}}{y}=\frac{{6}}{{7}}$.

Answer: $\frac{{4}}{{7}}$. Hint: either divide both sides by $\frac{{3}}{{2}}$ or multiple both sides by $\frac{{2}}{{3}}$.