Equations

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We discuss solving equations.

An equation is a statement expressing the equality between two quantities. This means that an equation will always have a single equals sign.

From Devyn’s question above, $\displaystyle \frac {74 + 84 + x}{3}$ is the average of the three exam scores, with the variable $x$ representing the third unknown exam score. What does the equation $\displaystyle \dfrac {74 + 84 + x}{3} = 82$ mean in this setting?

This is an equation, stating that the average of Devyn’s three exam scores is 82.

Usually, when dealing with an equation, we will be looking for values that make the equation true. A solution to an equation is a value that, when substituted in for the variables, yield a true number statement.

$\displaystyle \dfrac {74 + 84 + 88}{3} = 82$ , so $x = 88$ is a solution of the equation $\displaystyle \dfrac {74 + 84 + x}{3} = 82$ .

When we are asked to solve an equation, we are being asked to find all solutions. Exactly how to do that depends on the particular equation involved.

A linear equation in $x$ is an equation which is equivalent to one with the form $a x + b = 0$ , where $a$ and $b$ are constants, with $a \neq 0$ .

To solve a linear equation, we isolate the $x$ -term and divide by the coefficient.

Solve the linear equation $\displaystyle \dfrac {x}{3} - 9 = 4\left ( 2x + \dfrac {5}{2} \right )$ .

This doesn’t really look like $ax + b = 0$ when written like this. Let’s start by distributing the $4$ into the parentheses, then combining like terms. $\begin{align*} \frac{x}{3} - 9 &= 4\left( 2x + \dfrac{5}{2} \right) \\ \frac{x}{3} - 9 &= 8x + 10 \end{align*}$

That looks better. From here, we’ll move all the $x$ -terms to the right side of the equation, and all of the constant terms to the left.

$\begin{align*} - 9 - 10 &= 8x - \frac{x}{3}\\ -19 &= \frac{23}{3} x \\ x &= -19 \cdot \frac{3}{23} = -\frac{57}{23} \end{align*}$

Solve the linear equation $\displaystyle \dfrac {4}{5}\left (x+2\right ) - 3 = \dfrac {x-1}{2}$ .

It may be easier to clear fractions first.

x = $\answer {3}$ .