Multi-Step Linear Equations

Example 1. Solve the following equation: $$${5}{x}={3}{\left({2}{x}-{3}\right)}$$$.

This equation contains variables on both sides. We need to move all terms, containing variable, to one side, in order to isolate variable.

However, variable on the right side is inside parenthesis, so we need to expand parenthesis first:

$$${5}{x}={3}\cdot{2}{x}-{3}\cdot{3}$$$ (distributive property of multiplication: $$${a}{\left({b}-{c}\right)}={a}{b}-{a}{c}$$$)

$$${5}{x}={6}{x}-{9}$$$

Now, subtract $$${6}{x}$$$ from both sides of the equation:

$$${5}{x}{\color{red}{{-{6}{x}}}}={6}{x}-{9}{\color{red}{{-{6}{x}}}}$$$

$$${5}{x}-{6}{x}={6}{x}-{6}{x}-{9}$$$ (commutative property of addition: $$$-{9}-{6}{x}=-{6}{x}-{9}$$$)

$$${5}{x}-{6}{x}={0}-{9}$$$ (inverse property of addition: $$${6}{x}-{6}{x}={0}$$$)

$$${5}{x}-{6}{x}=-{9}$$$ (identity property of addition: $$${0}-{9}=-{9}$$$)

$$${\left({5}-{6}\right)}{x}=-{9}$$$ (distributive property of multiplicaton: $$${a}{c}-{b}{c}={\left({a}-{b}\right)}{c}$$$)

$$$-{x}=-{9}$$$

Finally, multiple both sides of equation by $$$-{1}$$$:

$$$-{x}\times{\left({\color{red}{{-{1}}}}\right)}={\left({\color{red}{{-{1}}}}\right)}$$$

$$${x}={9}$$$

So, $$${9}$$$ is root of the equation.

Example 2. Solve the following equation: $$${2}{\left({x}-{3}\right)}=\frac{{{10}{x}+{8}}}{{5}}$$$.

Rewrite equation a bit: $$${2}{\left({x}-{3}\right)}=\frac{{1}}{{5}}{\left({10}{x}+{8}\right)}$$$.

This equation contains variables on both sides. We need to move all terms, containing variable, to one side, in order to isolate variable.

But first we need to "free" variables out of parenthesis:

$$${2}{x}-{2}\cdot{3}=\frac{{1}}{{5}}\cdot{10}{x}+\frac{{1}}{{5}}\cdot{8}$$$ (distributive property of multiplication: $$${a}{\left({b}+{c}\right)}={a}{b}+{a}{c}$$$)

$$${2}{x}-{6}={2}{x}+\frac{{8}}{{5}}$$$

Now, subtract $$${2}{x}$$$ from both sides of the equation:

$$${2}{x}-{6}{\color{red}{{-{2}{x}}}}={2}{x}+\frac{{8}}{{5}}{\color{red}{{-{2}{x}}}}$$$

$$${2}{x}-{2}{x}-{6}={2}{x}-{2}{x}+\frac{{8}}{{5}}$$$ (commutative property of addition)

$$${0}-{6}={0}+\frac{{8}}{{5}}$$$ (inverse property of addition)

$$$-{6}\ne\frac{{8}}{{5}}$$$ (identity property of addition)

We've got incorrect equality, so this equation has no roots.

Example 3. Solve the following equation: $$$\frac{{1}}{{4}}{x}-\frac{{1}}{{2}}{x}+{5}=-\frac{{1}}{{4}}{x}+{5}$$$.

I will do this example a bit faster.

$$${\left(\frac{{1}}{{4}}-\frac{{1}}{{2}}\right)}{x}+{5}=-\frac{{1}}{{4}}{x}+{5}$$$

$$$-\frac{{1}}{{4}}{x}+{5}=-\frac{{1}}{{4}}{x}+{5}$$$

$$${5}={5}$$$

We've got correct equality, so this any number is root of this equation.

Example 4. Solve the following equation: $$$\frac{{\frac{{{2}{\left({x}-{2}\right)}}}{{3}}-{5}}}{{7}}=\frac{{1}}{{2}}{x}+{7}$$$.

Multiply both sides of equation by $$${7}$$$:

$$$\frac{{{2}{\left({x}-{2}\right)}}}{{3}}-{5}={7}{\left(\frac{{1}}{{2}}{x}+{7}\right)}$$$

$$$\frac{{{2}{\left({x}-{2}\right)}}}{{3}}-{5}={7}\cdot\frac{{1}}{{2}}{x}+{7}\cdot{7}$$$

$$$\frac{{{2}{\left({x}-{2}\right)}}}{{3}}-{5}=\frac{{7}}{{2}}{x}+{49}$$$

Add 5 to both sides of equation:

$$$\frac{{{2}{\left({x}-{2}\right)}}}{{3}}=\frac{{7}}{{2}}{x}+{54}$$$

Multiply both sides by $$${3}$$$:

$$${2}{\left({x}-{2}\right)}=\frac{{21}}{{2}}{x}+{162}$$$

Divide both sides by $$${2}$$$:

$$${x}-{2}=\frac{{21}}{{4}}{x}+{81}$$$

Add $$${2}$$$ to both sides of equation:

$$${x}=\frac{{21}}{{4}}{x}+{83}$$$

Subtract $$$\frac{{21}}{{4}}{x}$$$ from both sides of equation:

$$${x}-\frac{{21}}{{4}}{x}={83}$$$

$$$-\frac{{17}}{{4}}{x}={83}$$$

Divide both sides of equation by $$$-\frac{{17}}{{4}}$$$:

$$${x}=\frac{{83}}{{-\frac{{17}}{{4}}}}=-\frac{{332}}{{17}}$$$

So, $$$-\frac{{332}}{{17}}$$$ is root of the equation.