Literal equations Practice Questions (Grade 9)

Warm-upeasy

The simple interest formula is $I = Prt$ , where $I$ is interest, $P$ is principal, $r$ is rate, and $t$ is time. Which equation expresses $r$ in terms of $I$ , $P$ , and $t$ ?

The factors $P$ , $r$ , and $t$ are multiplied, so $r$ is isolated by using the inverse operation (division) on both sides of the equation $\dfrac{I}{\textcolor{red}{Pt}} = \dfrac{Prt}{\textcolor{red}{Pt}}$ . Simplifying gives $r = \dfrac{I}{Pt}$ .

Mid-gamemedium

The equation $y = mx + b$ is in slope-intercept form. Which equation expresses $x$ in terms of $y$ , $m$ , and $b$ ?

The term $b$ is added, so $mx$ is isolated by using the inverse operation (subtraction) on both sides, $y - \textcolor{red}{b} = mx$ . Then, $m$ and $x$ are multiplied, so $x$ is isolated by using the inverse operation (division) on both sides, $\dfrac{y - b}{\textcolor{red}{m}} = \dfrac{mx}{\textcolor{red}{m}}$ . Simplifying gives $x = \dfrac{y - b}{m}$ .

Mid-gamemedium

The perimeter formula for a rectangle is $P = 2l + 2w$ . Which equation expresses $w$ in terms of $P$ and $l$ ?

The term $2l$ is added, so $2w$ is isolated by using the inverse operation (subtraction) on both sides, $P - \textcolor{red}{2l} = 2w$ . Then, $2$ and $w$ are multiplied, so $w$ is isolated by using the inverse operation (division) on both sides, $\dfrac{P - 2l}{\textcolor{red}{2}} = \dfrac{2w}{\textcolor{red}{2}}$ . Simplifying gives $w = \dfrac{P - 2l}{2}$ .

Mid-gamemedium

The equation $ax + b = c$ relates values $a$ , $b$ , $c$ , and $x$ . Which equation expresses $x$ in terms of $a$ , $b$ , and $c$ ?

The term $b$ is added, so $ax$ is isolated by using the inverse operation (subtraction) on both sides, $c - \textcolor{red}{b} = ax$ . Then, $a$ and $x$ are multiplied, so $x$ is isolated by using the inverse operation (division) on both sides, $\dfrac{c - b}{\textcolor{red}{a}} = \dfrac{ax}{\textcolor{red}{a}}$ . Simplifying gives $x = \dfrac{c - b}{a}$ .

Mid-gamemedium

The volume formula for a rectangular prism is $V = lwh$ . Which equation expresses $h$ in terms of $V$ , $l$ , and $w$ ?

The factors $l$ , $w$ , and $h$ are multiplied, so $h$ is isolated by using the inverse operation (division) on both sides of the equation $\dfrac{V}{\textcolor{red}{lw}} = \dfrac{lwh}{\textcolor{red}{lw}}$ . Simplifying gives $h = \dfrac{V}{lw}$ .

Mid-gamemedium

The mean of $b$ and $c$ is $a$ , so $a = \dfrac{b + c}{2}$ . Which equation expresses $c$ in terms of $a$ and $b$ ?

The sum $b + c$ is divided by $2$ , so $b + c$ is isolated by using the inverse operation (multiplication) on both sides, $\textcolor{red}{2}a = b + c$ . Then, $b$ is added, so $c$ is isolated by using the inverse operation (subtraction) on both sides, $2a - \textcolor{red}{b} = c$ . Simplifying gives $c = 2a - b$ .

Mid-gamehard

The formula $F = \dfrac{9}{5}C + 32$ converts a Celsius temperature $C$ to Fahrenheit $F$ . Which equation expresses $C$ in terms of $F$ ?

The term $32$ is added, so $\dfrac{9}{5}C$ is isolated by using the inverse operation (subtraction) on both sides, $F - \textcolor{red}{32} = \dfrac{9}{5}C$ . Then, $C$ is multiplied by $\dfrac{9}{5}$ , so $C$ is isolated by using the inverse operation (multiplication by the reciprocal) on both sides, $\textcolor{red}{\dfrac{5}{9}}(F - 32) = \textcolor{red}{\dfrac{5}{9}} \cdot \dfrac{9}{5}C$ . Simplifying gives $C = \dfrac{5}{9}(F - 32)$ .

Buzzer beaterhard

The point-slope form of a line is $y - k = m(x - h)$ . Which equation expresses $x$ in terms of $y$ , $k$ , $m$ , and $h$ ?

The factor $m$ is multiplied by $(x - h)$ , so $x - h$ is isolated by using the inverse operation (division) on both sides, $\dfrac{y - k}{\textcolor{red}{m}} = x - h$ . Then, $h$ is subtracted from $x$ , so $x$ is isolated by using the inverse operation (addition) on both sides, $\dfrac{y - k}{m} + \textcolor{red}{h} = x$ . Simplifying gives $x = \dfrac{y - k}{m} + h$ .

Literal equations (SAT). Game on.

8 plays. No signup.

2 more. Played, not assigned.

Drill it inside a game.