5.3: Solving Equations of the Form ax = b and x/a = b
Equality Property of Division and Multiplication
Recalling that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side suggests the equality property of division and multiplication, which states:
- We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number, that is, if \(c \not = 0\), then \(a = b\) is equivalent to \(\dfrac{a}{c} = \dfrac{b}{c}\).
- We can obtain an equivalent equation by multiplying both sides of the equation by the same nonzero number, that is, if \(c \not = 0\), then \(a = b\) is equivalent to \(ac=bc\).
We can use these results to isolate x, thus solving the equation for x.
Solving \(ax = b\) for \(x\)
\(\begin{array}{flushleft}
ax&=&b&a\text{ is associated with } x \text{ by multiplication. }\\
&&&\text{Undo the association by diving both sides by } a\\
\dfrac{ax}{a}&=&\dfrac{b}{a}\\
\dfrac{\not{a}x}{a}&=&\dfrac{b}{a}\\
1 \cdot x &=&\dfrac{b}{a}&\dfrac{a}{a}=1\text{ and }1 \text{ is the multiplicative identity. } 1 \cdot x = x
\end{array}\)
Solving \(\dfrac{x}{a} = b\) for \(x\)
\(\begin{array}{flushleft}
x&=&\dfrac{b}{a}&\text{This equation is equivalent to the first and is solved by } x\\
\dfrac{x}{a}&=&b&a\text{ is associated with } x \text{ by division. Undo the association }\\
&&&\text{by multiplying both sides by } a\\
a \cdot \dfrac{x}{a}&=&a \cdot b\\
\not{a} \cdot \dfrac{x}{\not{a}}&=&ab\\
1 \cdot x&=&ab&\dfrac{a}{a}=1\text{ and } 1 \text{ is the multiplicative identity. } 1 \cdot x = x\\
x&=&ab&\text{ This equation is equivalent to the first and is solved for } x
\end{array}\)
Solving \(ax=b\) and \(\dfrac{x}{a} = b\) for \(x\)
To solve \(ax = b\) for \(x\), divide both sides of the equation by \(a\).
To solve \(\dfrac{x}{a} = b\) for \(x\), multiply both sides of the equation by \(a\).
Sample Set A
Solve \(5x = 35\) for \(x\).
\(\begin{array}{flushleft}
5x&=&35&5\text{ is associated with } x \text{ by multiplication. Undo the}\\
&&&\text{association by dividing both sides by } 5.\\
\dfrac{5x}{5}&=&\dfrac{35}{5}\\
\dfrac{\not{5}x}{\not{5}}&=&7\\
1 \cdot x&=&7&\dfrac{5}{5}=1\text{ and }1 \text{ is multiplicative identity. } 1 \cdot x = x.\\
x&=&7
\end{array}\)
Check:
\(\begin{array}{flushleft}
5(7)&=&35&\text{ Is this correct? }\\
35&=&35&\text{ Yes, this is correct. }
\end{array}\)
Solve \(\dfrac{x}{4} = 5\) for \(x\).
\(\begin{array}{flushleft}
\dfrac{x}{4}&=&5&4\text{ is associated with } x \text{ by division. Undo the association by}\\
&&&\text{multiplying both sides by } 4.\\
4 \cdot \dfrac{x}{4}&=&4 \cdot 5\\
\not{4} \cdot \dfrac{x}{\not{4}}&=&4 \cdot 5\\
1 \cdot x&=&20&\dfrac{4}{4}=1\text{ and } 1 \text{ is the multiplicative identity. } 1 \cdot x = x.\\
x&=&20
\end{array}\)
Check:
\(\begin{array}{flushleft}
\dfrac{20}{4}&=&5&\text{ Is this correct? }\\
5&=&5&\text{ Yes, this is correct.}
\end{array}\)
Solve \(\dfrac{2y}{9} = 3\) for \(y\).
Method (1) (Use of canceling):
\(\begin{array}{flushleft}
\dfrac{2y}{9}&=&3&9\text{ is associated with } y \text{ by division. Undo the association by}\\
&&&\text{multiplying both sides by } 9.\\
(\not{9})(\dfrac{2y}{not{9}})&=&(9)(3)\\
2y&=&27&2\text{ is associated with } y { by multiplication. Undo the}\\
&&&\text{association by dividing both sides by } 2.\\
\dfrac{not{2}y}{not{2}}&=&\dfrac{27}{2}\\
y&=&\dfrac{27}{2}
\end{array}\)
Check:
\(\begin{array}{flushleft}
\dfrac{\not{2}(\dfrac{27}{\not{2}})}{9}&=&3&\text{ Is this correct?}\\
\dfrac{27}{9}&=&3&\text{ Is this correct?}\\
3&=&3&\text{ Yes, this is correct. }
\end{array}\)
Method (2)(Use of reciprocals):
\(\begin{array}{flushleft}
\dfrac{2y}{9}&=&3&\text{Since } \dfrac{2y}{9}=\dfrac{2}{9}y, \dfrac{2}{9} \text{ is associated with } y \text{ by multiplication. }\\
&&&\text{Then, Since } \dfrac{9}{2} \cdot \dfrac{2}{9}=1\text{, the multiplicative identity, we can }\\
&&&\text{undo the associative by multiplying both sides by } \dfrac{9}{2}\\
(\dfrac{9}{2})(\dfrac{2y}{9})&=&(\dfrac{9}{2})(3)\\
(\dfrac{9}{2} \cdot \dfrac{2}{9})y&=&\dfrac{27}{2}\\
1 \cdot y&=&\dfrac{27}{2}\\
y&=&\dfrac{27}{2}
\end{array}\)
Solve the literal equation \(\dfrac{4ax}{m} = 3b\) for \(x\).
\(\begin{array}{flushleft}
\dfrac{4ax}{m}&=&3b&m\text{ is associated with } x \text{ by division. Undo the association by }\\
&&&\text{multiplying both sides by } m.\\
\not{m}(\dfrac{4ax}{\not{m}})&=&m \cdot 3b\\
4ax&=&3bm&4a\text{ is associated with } x \text{ by multiplication. Undo the }\\
&&&\text{association by multiplying both sides by } 4a\\
\dfrac{\not{4a}x}{\not{4a}}&=&\dfrac{3bm}{4a}\\
x&=&\dfrac{3bm}{4a}
\end{array}\)
Check:
\(\begin{array}{flushleft}
\dfrac{4a(\dfrac{3bm}{4a})}{m}&=&3b&\text{ Is this correct? }\\
\dfrac{\not{4a}(\dfrac{3bm}{\not{4a}})}{m}&=&3b&\text{ Is this correct?}\\
\dfrac{3b\not{m}}{\not{m}}&=&3b&\text{ Is this correct?}\\
3b&=&3b&\text{ Yes, this is correct.}
\end{array}\)
Practice Set A
Solve \(6a=42\) for \(a\).
- Answer
-
\(a = 7\)
Solve \(−12m=16\) for \(m\).
- Answer
-
\(m = -\dfrac{4}{3}\)
Solve \(\dfrac{y}{8} = -2\) for \(y\)
- Answer
-
\(y = -16\)
Solve \(6.42x = 1.09\) for \(x\)
- Answer
-
\(x = 0.17\) (rounded to two decimal places)
Solve \(\dfrac{5k}{12} = 2\) for \(k\).
- Answer
-
\(k = \dfrac{24}{5}\)
Solve \(\dfrac{-ab}{2c} = 4d\) for \(b\).
- Answer
-
\(b = \dfrac{-8cd}{a}\)
Solve \(\dfrac{3xy}{4} = 9xh\) for \(y\).
- Answer
-
\(y = 12h\)
Solve \(\dfrac{2k^2mn}{5pq} = -6n\) for \(m\).
- Answer
-
\(m = \dfrac{-15pq}{k^2}\)
Exercises
In the following problems, solve each of the conditional equations.
\(3x = 42\)
- Answer
-
\(x = 14\)
\(5y = 75\)
\(6x = 48\)
- Answer
-
\(x=8\)
\(8x = 56\)
\(4x = 56\)
- Answer
-
\(x=14\)
\(3x = 93\)
\(5a = −80\)
- Answer
-
\(a=−16\)
\(9m = −108\)
\(6p = −108\)
- Answer
-
\(p=−18\)
\(12q = −180\)
\(−4a = 16\)
- Answer
-
\(a=−4\)
\(−20x = 100\)
\(−6x = −42\)
- Answer
-
\(x=7\)
\(−8m = −40\)
\(−3k = 126\)
- Answer
-
\(k=−42\)
\(−9y = 126\)
\(\dfrac{x}{6} = 1\)
- Answer
-
\(x=6\)
\(\dfrac{a}{5} = 6\)
\(\dfrac{k}{7} = 6\)
- Answer
-
\(k=42\)
\(\dfrac{x}{3} = 72\)
\(\dfrac{x}{8} = 96\)
- Answer
-
\(x = 768\)
\(\dfrac{y}{-3} = -4\)
\(\dfrac{m}{7} = -8\)
- Answer
-
\(m = -56\)
\(\dfrac{k}{18} = 47\)
\(\dfrac{f}{-62} = 103\)
- Answer
-
\(f = -6386\)
\(3.06m= 12.546\)
\(5.012k = 0.30072\)
- Answer
-
\(k=0.06\)
\(\dfrac{x}{2.19} = 5\)
\(\dfrac{y}{4.11} = 2.3\)
- Answer
-
\(y=9.453\)
\(\dfrac{4y}{7} = 2\)
\(\dfrac{3m}{10} = -1\)
- Answer
-
\(m = \dfrac{-10}{3}\)
\(\dfrac{5k}{6} = 8\)
\(\dfrac{8h}{-7} = -3\)
- Answer
-
\(h = \dfrac{21}{8}\)
\(\dfrac{-16z}{21} = -4\)
Solve \(pq = 7r\) for \(p\)
- Answer
-
\(p = \dfrac{7r}{q}\)
Solve \(m^2n = 2s\) for \(n\)
Solve \(2.8ab = 5.6d\) for \(b\)
- Answer
-
\(b = \dfrac{2d}{a}\)
Solve \(\dfrac{mnp}{2k} = 4k\) for \(p\)
Solve \(\dfrac{-8a^2b}{3c} = -5a^2\) for \(b\).
- Answer
-
\(b = \dfrac{15c}{8}\)
Solve \(\dfrac{3pcb}{2m} = 2b\) for \(pc\)
Solve \(\dfrac{8rst}{3p} = -2prs\) for \(t\).
- Answer
-
\(t = -\dfrac{-3p^2}{4}\)
Exercises for Review
Simplify \((\dfrac{2x^0y^0z^3}{z^2})^5\)
Classify \(10x^3-7x\) as a monomial, binomial, or trinomial. State its degree and write the numerical coefficient of each item.
- Answer
-
binomial; 3rd degree; 10,−7
Simplify \(3a^2-2a+4a(a+2)\)
Specify the domain of the equation \(y = \dfrac{3}{7+x}\).
- Answer
-
all real numbers except −7
Solve the conditional equation \(x+6=−2\).