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.

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.

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.

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.