A literal equation is synonymous with a formula and similar to solving general linear equations because we apply the same method. We say, methods never change, just the problems. The only difference is we have several variables in the equation and we will attempt to solve for one specific variable of the formula. For example, we may have a formula such as \(A = πr^2 +πrs\), the formula for surface area of a right circular cone, and we may be interested in solving for the variable \(s\). This means we want to isolate the variable \(s\) so the equation has s isolated on one side, and everything else on the other. This looks like \[s=\frac{A-\pi r^2}{\pi r}\nonumber\]
This second equation gives the same information as the first, meaning they are algebraically equivalent. However, the original formula gives area, while the other gives \(s\), the slant height of the cone. In this section, we discuss the process in which we start from the first equation and result in the second equation.
Example \(\PageIndex{1}\)
Let’s take a look at these two examples below, side by side. The left equation is a familiar one-step equation and the right equation is also a one-step equation, this time a literal equation (or formula).
\[\begin{array}{rcl} 3x=12 & wx=z &\text{Both have coefficients} \\ &&\text{Multiply by the reciprocal of }3\text{ and }w\text{, respectively} \\ \color{blue}{\frac{1}{3}}\color{black}{}\cdot 3x=\color{blue}{\frac{1}{3}}\color{black}{}\cdot 12 & \color{blue}{\frac{1}{w}}\color{black}{}\cdot wx=z\cdot\color{blue}{\frac{1}{w}}\color{black}{}&\text{Simplify} \\ x=4& x=\frac{z}{w}&\text{Solution}\end{array}\nonumber\]
We used the same process for solving \(3x = 12\) for \(x\) as we did for solving \(wx = z\) for \(x\). Because we are solving for \(x\), we treat all the other variables the same way we would treat numbers or coefficients. Thus, we applied the multiplication property and multiplied by the reciprocal of \(3\) and \(w\) to isolate \(x\).
Solving for a Variable with One and Two-Step Equations
Example \(\PageIndex{2}\)
Solve the equation \(m + n = p\) for \(n\).
Solution
\[\begin{array}{rl} m+n=p&\text{Add the opposite of }m \\ m+n+\color{blue}{(-m)}\color{black}{}=p+\color{blue}{(-m)}\color{black}{}&\text{Simplify} \\ n=p-m&\text{Solution}\end{array}\nonumber\]
Since \(p\) and \(m\) are not like terms, they cannot be combined. Hence, \(n = p − m\).
Example \(\PageIndex{3}\)
Solve the equation \(a(x − y) = b\) for \(x\).
Solution
\[\begin{array}{rl} a(x-y)=b&\text{Distribute} \\ ax-ay=b&\text{Add the opposite of }ay \\ ax+ay+\color{blue}{(-ay)}\color{black}{}=b+\color{blue}{(-ay)}\color{black}{}&\text{Simplify} \\ ax=b-ay &\text{Isolate }x\text{ by multiplying by the reciprocal of }a \\ \color{blue}{\frac{1}{a}}\color{black}{}\cdot ax=(b-ay)\cdot\color{blue}{\frac{1}{a}}\color{black}{}&\text{Simplify} \\ x=\frac{b-ay}{a}&\text{Solution}\end{array}\nonumber\]
Equivalently, \(x\) can be written as \(\frac{b}{a} − y\) by simplifying the fraction. However, it is common practice to leave it as one fraction.
Example \(\PageIndex{4}\)
Solve the equation \(y = mx + b\) for \(m\).
Solution
\[\begin{array}{rl}y=mx+b&\text{Isolate the variable term by adding the opposite of }b \\ y+\color{blue}{(-b)}\color{black}{}=mx+b+\color{blue}{(-b)}\color{black}{}&\text{Simplify} \\ y-b=mx&\text{Isolate }m\text{ by multiplying by the reciprocal of }x \\ \color{blue}{\frac{1}{x}}\color{black}{}\cdot (y-b)=mx\cdot\color{blue}{\frac{1}{x}}\color{black}{}&\text{Simplify} \\ \frac{y-b}{x}=m&\text{Rewrite with }m\text{ on the left side} \\ m=\frac{y-b}{x}&\text{Solution}\end{array}\nonumber\]
Solving for a Variable in Multiple Steps
Example \(\PageIndex{5}\)
Solve the equation \(A = πr^2 + πrs\) for \(s\). This should remind you of the equation in the beginning of the section.
Solution
\[\begin{array}{rl}A=\pi r^2+\pi rs&\text{Isolate the variable term by adding the opposite of }\pi r^2 \\ A+\color{blue}{(-\pi r^2)}\color{black}{}=\pi r^2+\pi rs+\color{blue}{(-\pi r^2)}\color{black}{}&\text{Simplify} \\ A-\pi r^2=\pi rs&\text{Isolate }s\text{ by multiplying by the reciprocal of }\pi r \\ \color{blue}{\frac{1}{\pi r}}\color{black}{}\cdot(A-\pi r^2)=\pi rs\cdot\color{blue}{\frac{1}{\pi r}}\color{black}{}&\text{Simplify} \\ \frac{A-\pi r^2}{\pi r}=s&\text{Rewrite with }s\text{ on the left side} \\ s=\frac{A-\pi r^2}{\pi r}&\text{Solution}\end{array}\nonumber\]
Solving for a Variable with Fractions
Formulas often include fractions and we can solve with the same method as used previously. First, identify the LCD, and then multiply each term by the LCD. After we clear denominators, we obtain a general equation and solve as usual.
Example \(\PageIndex{6}\)
Solve the equation \(h = \frac{2m}{n}\) for \(m\).
Solution
\[\begin{array}{rl}h=\frac{2m}{n}&\text{Multiply by the LCD=}n \\ \color{blue}{n}\color{black}{}\cdot h=\frac{2m}{n}\cdot\color{blue}{n}\color{black}{}&\text{Simplify} \\ nh=2m&\text{Multiply by the reciprocal of }2 \\ \color{blue}{\frac{1}{2}}\color{black}{}\cdot nh=2m\cdot\color{blue}{\frac{1}{2}}\color{black}{}&\text{Simplify} \\ \frac{nh}{2}=m&\text{Rewrite with }m\text{ on the left side} \\ m=\frac{nh}{2}&\text{Solution}\end{array}\nonumber\]
Example \(\PageIndex{7}\)
Solve the equation \(\frac{a}{b}+\frac{c}{b}=e\) for \(a\).
Solution
\[\begin{array}{rl}\frac{a}{b}+\frac{c}{b}=e&\text{Multiply each term by the LCD=}b \\ \color{blue}{b}\color{black}{}\cdot\frac{a}{b}+\color{blue}{b}\color{black}{}\cdot\frac{c}{b}=e\cdot\color{blue}{b}\color{black}{}&\text{Simplify} \\ a+c=eb&\text{Add the opposite of }c \\ a+c+\color{blue}{(-c)}\color{black}{}=eb+\color{blue}{(-c)}\color{black}{}&\text{Simplify} \\ a=eb-c&\text{Solution}\end{array}\nonumber\]
Example \(\PageIndex{8}\)
Solve the equation \(a=\frac{A}{2-b}\) for \(b\).
Solution
\[\begin{array}{rl}a=\frac{A}{2-b}&\text{Multiply each term by the LCD}=(2-b) \\ \color{blue}{(2-b)}\color{black}{}\cdot a=\frac{A}{2-b}\cdot\color{blue}{(2-b)}\color{black}{}&\text{Simplify} \\ a(2-b)=A&\text{Distribute} \\ 2a-2b=A&\text{Isolate the variable term by adding the opposite of }2a \\ 2a-2b+\color{blue}{(-2a)}\color{black}{}=A+\color{blue}{(-2a)}\color{black}{}&\text{Simplify} \\ -2b=A-2a&\text{Multiply by the reciprocal of }-2 \\ \color{blue}{-\frac{1}{2}}\color{black}{}\cdot -2b=(A-2a)\cdot\color{blue}{-\frac{1}{2}}\color{black}{}&\text{Simplify} \\ b=-\frac{(A-2a)}{2}&\text{Distribute the negative} \\ b=\frac{-A+2a}{2}&\text{Solution}\end{array}\nonumber\]
Note, we could also write the solution as \(b = \frac{2a − A}{2}\), where the positive term is written first in the numerator. It’s not necessary, but for aesthetic reasons, we can write \(b\) this way.
Note
The father of algebra, Persian mathematician, Muhammad ibn Musa Khwarizmi, introduced the fundamental idea of balancing by subtracting the same term from the other side of the equation. He called this process al-jabr, which later became the world Algebra.
Literal Equations Homework
Solve each of the following equations for the indicated variable.
Exercise \(\PageIndex{1}\)
\(ab = c\) for \(b\)
Exercise \(\PageIndex{2}\)
\(\frac{f}{g}x = b\) for \(x\)
Exercise \(\PageIndex{3}\)
\(3x = \frac{a}{b}\) for \(x\)
Exercise \(\PageIndex{4}\)
\(E = mc^2\) for \(m\)
Exercise \(\PageIndex{5}\)
\(V = \frac{4}{3} πr^3\) for \(π\)
Exercise \(\PageIndex{6}\)
\(a + c = b\) for \(c\)
Exercise \(\PageIndex{7}\)
\(c = \frac{4y}{m + n}\) for \(y\)
Exercise \(\PageIndex{8}\)
\(V = \frac{πDn}{12}\) for \(D\)
Exercise \(\PageIndex{9}\)
\(P = n(p − c)\) for \(n\)
Exercise \(\PageIndex{10}\)
\(T = \frac{D − d}{L}\) for \(D\)
Exercise \(\PageIndex{11}\)
\(L = L_0(1 + at)\) for \(L_0\)
Exercise \(\PageIndex{12}\)
\(2m + p = 4m + q\) for \(m\)
Exercise \(\PageIndex{13}\)
\(\frac{k − m}{r} = q\) for \(k\)
Exercise \(\PageIndex{14}\)
\(h = vt − 16t^2\) for \(v\)
Exercise \(\PageIndex{15}\)
\(Q_1 = P(Q_2 − Q_1)\) for \(Q_2\)
Exercise \(\PageIndex{16}\)
\(R = \frac{kA(T_1 + T_2)}{d}\) for \(T_1\)
Exercise \(\PageIndex{17}\)
\(ax + b = c\) for \(a\)
Exercise \(\PageIndex{18}\)
\(lwh = V\) for \(w\)
Exercise \(\PageIndex{19}\)
\(\frac{1}{a} + b = \frac{c}{a}\) for \(a\)
Exercise \(\PageIndex{20}\)
\(at − bw = s\) for \(t\)
Exercise \(\PageIndex{21}\)
\(ax + bx = c\) for \(a\)
Exercise \(\PageIndex{22}\)
\(x + 5y = 3\) for \(y\)
Exercise \(\PageIndex{23}\)
\(3x + 2y = 7\) for \(y\)
Exercise \(\PageIndex{24}\)
\(5a − 7b = 4\) for \(b\)
Exercise \(\PageIndex{25}\)
\(4x − 5y = 8\) for \(y\)
Exercise \(\PageIndex{26}\)
\(g = \frac{h}{i}\) for \(h\)
Exercise \(\PageIndex{27}\)
\(p = \frac{3y}{q}\) for \(y\)
Exercise \(\PageIndex{28}\)
\(\frac{ym}{b} = \frac{c}{d}\) for \(y\)
Exercise \(\PageIndex{29}\)
\(DS = ds\) for \(D\)
Exercise \(\PageIndex{30}\)
\(E = \frac{mv^2}{2}\) for \(m\)
Exercise \(\PageIndex{31}\)
\(x − f = g\) for \(x\)
Exercise \(\PageIndex{32}\)
\(\frac{rs}{a − 3} = k\) for \(r\)
Exercise \(\PageIndex{33}\)
\(F = k(R − L)\) for \(k\)
Exercise \(\PageIndex{34}\)
\(S = L + 2B\) for \(L\)
Exercise \(\PageIndex{35}\)
\(I = \frac{E_a − E_q}{R}\) for \(E_a\)
Exercise \(\PageIndex{36}\)
\(ax + b = c\) for \(x\)
Exercise \(\PageIndex{37}\)
\(q = 6(L − p)\) for \(L\)
Exercise \(\PageIndex{38}\)
\(R = aT + b\) for \(T\)
Exercise \(\PageIndex{39}\)
\(S = πrh + πr^2\) for \(h\)
Exercise \(\PageIndex{40}\)
\(L = π(r_1 + r_2) + 2d\) for \(r_1\)
Exercise \(\PageIndex{41}\)
\(P = \frac{V_1(V_2 − V_1)}{g}\) for \(V_2\)
Exercise \(\PageIndex{42}\)
\(rt = d\) for \(r\)
Exercise \(\PageIndex{43}\)
\(V = \frac{πr^2h}{3}\) for \(h\)
Exercise \(\PageIndex{44}\)
\(\frac{1}{a} + b = \frac{c}{a}\) for \(b\)
Exercise \(\PageIndex{45}\)
\(at − bw = s\) for \(w\)
Exercise \(\PageIndex{46}\)
\(x + 5y = 3\) for \(x\)
Exercise \(\PageIndex{47}\)
\(3x + 2y = 7\) for \(x\)
Exercise \(\PageIndex{48}\)
\(5a − 7b = 4\) for \(a\)
Exercise \(\PageIndex{49}\)
\(4x − 5y = 8\) for \(x\)
Exercise \(\PageIndex{50}\)
\(C = \frac{5}{9} (F − 32)\) for \(F\)
