# Two-Step Linear Equations

Two-step linear equation is an equation, that requires two actions (operations) in order to be solved.

Basically, there are 2 kinds of such equations:

• ${2}{x}+{5}={15}$ (add/subtract, then multiply/divide)
• $\frac{{{a}-{2}}}{{4}}={3}$ (multiply/divide, then add/subtract)

Above 2 kinds of linear equations are general. Addition/subtraction, multiplication/division can come in any order. Luckily, you don't need to remember all possible situations. 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 two-step linear equations.

Example 1. Solve the following equation: ${2}{x}-{3}={5}$.

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 adding 3 to both sides of equation:

${2}{x}-{3}{\color{red}{{+{3}}}}={5}{\color{red}{{+{3}}}}$

${2}{x}+{0}={8}$ (inverse property of addition: $-{3}+{3}={0}$)

${2}{x}={8}$ (identity property of addition: ${2}{x}+{0}={2}{x}$)

This was the first step.

Second step is the following: in order to get rid of 2 near ${x}$, we divide both sides of equation by 2.

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

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

${x}={4}$ (idenity property of multiplication: ${1}\times{x}={x}$)

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

Next, let's solve an example, when we need to perform multiplication first.

Example 2. Solve the following equation: $\frac{{{a}+{3}}}{{2}}={7}$.

First step is to multiply both sides of equation by 2.

${\color{red}{{{2}\times}}}\frac{{{a}+{3}}}{{2}}={\color{red}{{{2}\times}}}{7}$

${1}\times{\left({a}+{3}\right)}={14}$ (inverse property of multiplication: ${2}\times\frac{{1}}{{2}}={1}$)

${a}+{3}={14}$ (identity property of multiplication: ${1}\times{\left({a}+{3}\right)}={a}+{3}$)

Note, that it doesn't matter whether we write ${2}\times\frac{{{a}+{3}}}{{2}}$ or $\frac{{{a}+{3}}}{{2}}\times{2}$ due to the commutative property of multiplication. Same applies to ${2}\times{7}$ (we could write it as ${7}\times{2}$).

Second step is the following: in order to get rid of 3, we subtract 3 from both sides of equation.

${a}+{3}{\color{red}{{-{3}}}}={14}{\color{red}{{-{3}}}}$

${a}+{0}={11}$ (inverse property of addition: ${3}-{3}={3}+{\left(-{3}\right)}={0}$)

${a}={11}$ (identity property of addition: ${a}+{0}={a}$)

So, ${11}$ is root of the equation.

One more example.

Example 3. Solve the following equation: ${3}{\left({y}-{5}\right)}={0}$.

First step is to divide both sides of equation by 3.

$\frac{{{3}{\left({y}-{5}\right)}}}{{{\color{red}{{{3}}}}}}=\frac{{0}}{{{\color{red}{{{3}}}}}}$

${1}\times{\left({y}-{5}\right)}={0}$ (inverse property of multiplication: $\frac{{3}}{{3}}={3}\times\frac{{1}}{{3}}={1}$)

${y}-{5}={0}$ (identity property of multiplication: ${1}\times{\left({y}-{5}\right)}={y}-{5}$)

Second step is the following: in order to get rid of $-{5}$, we add 5 from both sides of equation.

${y}-{5}{\color{red}{{+{5}}}}={0}{\color{red}{{+{5}}}}$

${y}+{0}={5}$ (inverse property of addition: $-{5}+{5}={0}$)

${y}={5}$ (identity property of addition: ${y}+{0}={y}$)

So, ${5}$ is root of the equation.

Finally, consider example, when we need to combine like terms.

Example 4. Solve the following equation: ${5}{x}-{2}{x}={12}$.

We have two terms with variables, so need to combine them.

${\left({5}-{2}\right)}{x}={12}$ (distributive property of multiplication: ${a}{x}-{b}{x}={\left({a}-{b}\right)}{x}$)

${3}{x}={12}$

Second step is the following: in order to get rid of ${3}$ near ${x}$, we divide both sides of equation by ${3}$.

$\frac{{{3}{x}}}{{{\color{red}{{{3}}}}}}=\frac{{12}}{{{\color{red}{{{3}}}}}}$

${1}\times{x}={4}$ (inverse property of multiplication: $\frac{{3}}{{3}}=\frac{{1}}{{3}}\times{3}={1}$)

${x}={4}$ (identity property of multiplication: ${1}\times{x}={x}$)

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

Now, it is time to exercise.

Exercise 1. Solve the equation ${3}{x}-{3}={12}$.

Answer: ${5}$.

Exercise 2. Find roots of the equation $\frac{{x}}{{2}}+{3}={2}$.

Answer: $-{2}$.

Exercise 3. Solve the equation ${3}{\left({y}+\frac{{7}}{{2}}\right)}=\frac{{2}}{{5}}$.

Answer: $-\frac{{101}}{{30}}$.

Exercise 4. Solve the equation $\frac{{1}}{{3}}{a}-{2}{a}={5}$.

Answer: $-{3}$.

Exercise 5. Solve the equation $\frac{{{x}-\frac{{1}}{{2}}}}{{4}}={2}$.

Answer: $\frac{{17}}{{2}}={8.5}$.