One-Step Linear Equations

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.

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 $-\frac{{7}}{{30}}\times{15}$).