4.4 Beyond the Basics
Introduction
In this section, we will introduce three types of equations that are just a little bit different than what we've seen before.
These new equations can easily be confused for one of the earlier types, so we will have to proceed with caution to identify them correctly.
Equations with Negative Coefficients
When solving an equation for a variable like \(x\), we sometimes run into the expression \(-x\).
Since \(-x\) can be viewed as \(-1\cdot x\), if we are able to isolate \(-x\), then both sides could be divide by \(-1\), to give us the value for positive \(x\).
When solving an equation where the variable is negative, there are a few common errors that should be avoided. The first error is copying the equation incorrectly. For instance, \(2-x=9\) is not the same as \(x-2=9\).
Solve \(2 - x = 9\) for \(x\).
Solution \begin{align} 2 - x &= 9\ && \small \color{#5fa2ce} {\text{Original equation}} \\ -x &= 7 && \small \color{#5fa2ce} {\text{Subtract }2} \\ x &= -7 && \small \color{#5fa2ce} {\text{Divide by }-1} \end{align}
The second error is not following the proper order of operations. For example, it is very tempting to simplify \(5-2x\) as \(3x\), but that places subtraction before multiplication, which is not the correct order.
Solve \(5 - 2x = 21\) for \(x\).
Solution \begin{align} 5 - 2x &= 21 && \small \color{#5fa2ce} {\text{Given equation}} \\ -2x &= 16 && \small \color{#5fa2ce} {\text{Subtract }5} \\ x &= -8 && \small \color{#5fa2ce} {\text{Divide by }-2} \end{align}
Some prefer to rewrite the left side of the equation as \(-2x+5\) to make the operations easier to see.
\begin{align} -2x + 5 &= 21 && \small \color{#5fa2ce} {\text{Rewritten equation}} \\ -2x &= 16 && \small \color{#5fa2ce} {\text{Subtract }5} \\ x &= -8 && \small \color{#5fa2ce} {\text{Divide by }-2} \end{align}Both produce the same result.
This same error is also possible in situations where distribution could be used. In the next example we will solve \(7-3(x-1)=34\). Note that the \(7\) and the \(-3\) cannot be combined since the \(-3\) is multiplying \((x-1)\) and \(7\) is not.
Solve \(7-3(x-1)=34\) for \(x\).
Solution \begin{align} 7-3(x-1) &= 34 && \small \color{#5fa2ce} {\text{Given equation}} \\ -3(x-1) &= 27 && \small \color{#5fa2ce} {\text{Subtract }7} \\ x -1 &= -9 && \small \color{#5fa2ce} {\text{Divide by }-3} \\ x &= -8 && \small \color{#5fa2ce} {\text{Add }1} \end{align}
Or, if you prefer to use distribution:
\begin{align} 7-3(x-1) &= 34 && \small \color{#5fa2ce} {\text{Given equation}} \\ 7-3x+3 &= 34 && \small \color{#5fa2ce} {\text{Distribute }-3} \\ 10-3x &= 34 && \small \color{#5fa2ce} {\text{Combine constants}} \\ -3x &= 24 && \small \color{#5fa2ce} {\text{Subtract }10} \\ x &= -8 && \small \color{#5fa2ce} {\text{Divide by }-3} \end{align}Equations with a Variable in the Denominator
The next type of equation to watch out for is one where the variable is in the denominator. It's critical to recognize that dividing something by \(x\) is not the same as dividing \(x\) by something. For instance, \(\frac{5}{x}\) is not the same as \(\frac{x}{5}\).
Luckily, equations with a variable in the denominator can usually be converted into ones where the variable isn't in the denominator.
This is done by "clearing the fractions" by multiplying each term by the least common denominator (LCD). We did a similar step to simplify equations in section 4.1.
Solve \(\frac{5}{x} = 8\) for \(x\).
Solution
Since the denominators are \(x\) and \(1\), the LCD is \(x\).
\begin{align} \frac{5}{x} &= 8 && \small \color{#5fa2ce} {\text{Given equation}} \\ x\cdot \frac{5}{x} &= x\cdot 8 && \small \color{#5fa2ce} {\text{Multiply by }x} \\ 5 &= 8x && \small \color{#5fa2ce} {\text{Simplfy}} \\ x &= \frac{5}{8} && \small \color{#5fa2ce} {\text{Divide by }8} \end{align}Equations where both sides are fractions are known as proportions. With proportions, the process of clearing the fractions can be viewed as cross multiplication.
To cross multiply fractions, multiply the top of each fraction with the bottom of the other fraction. This converts an equation like \(\frac{a}{b}=\frac{c}{d}\) into one of the form \(a\cdot d = b\cdot c\).
Solve \(\frac{10}{x} = \frac{2}{7}\) for \(x\).
Solution \begin{align} \frac{10}{x} &= \frac{2}{7} && \small \color{#5fa2ce} {\text{Given equation}} \\ 10 \cdot 7 &= 2 \cdot x && \small \color{#5fa2ce} {\text{cross multiply to eliminate the denominator}} \\ 70 &= 2x && \small \color{#5fa2ce} {\text{Simplify}} \\ 35 &= x && \small \color{#5fa2ce} {\text{Divide by }2} \end{align}
Sometimes cross multiplication will lead to situations where distribution and/or combining like terms is necessary.
Solve \(\frac{3}{x+1} = \frac{4}{3x}\) for \(x\).
Solution \begin{align} \frac{3}{x+1} &= \frac{4}{3x} && \small \color{#5fa2ce} {\text{Given equation}} \\ 9x &= 4(x+1) && \small \color{#5fa2ce} {\text{cross multiply to eliminate the denominator}} \\ 9x &= 4x + 4 && \small \color{#5fa2ce} {\text{Distribute }4} \\ 5x &= 4 && \small \color{#5fa2ce} {\text{Subtract }4x} \\ x &= \frac{4}{5} && \small \color{#5fa2ce} {\text{Divide by }5} \end{align}
Cross multiplication is a handy tool, but don't forget that it can only be used when both sides are single fractions. If one side has two or more terms then we cannot use cross multiplication and need to fall back on the method of multiplying by a common denominator.
Solve \(\frac{4}{x} + 3 = \frac{23}{5x}\) for \(x\).
Solution
Since the left side is not a single fraction, we cannot cross multiply. Instead, we'll multiply by the common denominator, which is \(5x\).
\begin{align} \frac{4}{x} + 3 &= \frac{23}{5x} && \small \color{#5fa2ce} {\text{Given equation}} \\ 5x\left(\frac{4}{x} + 3\right) &= 5x \cdot \frac{23}{5x} && \small \color{#5fa2ce} {\text{Multiply both sides by } 5x} \\ 5x \cdot \frac{4}{x} + 5x\cdot 3 &= 5x \cdot \frac{23}{5x} && \small \color{#5fa2ce} {\text{On the left, distribute } 5x} \\ \frac{20x}{x} + 15x &= \frac{115x}{5x} && \small \color{#5fa2ce} {\text{Multiply}} \\ 20 + 15x &= 23 && \small \color{#5fa2ce} {\text{Simplify fractions}} \\ 15x &= -3 && \small \color{#5fa2ce} {\text{Subtract 20 from both sides}} \\ x &= \frac{-3}{15} && \small \color{#5fa2ce} {\text{Divide both sides by }15} \\ x &= \frac{-1}{5} && \small \color{#5fa2ce} {\text{Reduce fraction} x} \end{align}Equations with a Variable in the Exponent
This final type of equation might look familiar, and that's the challenge. It takes a careful eye to notice that expressions like \(x^2\) and \(2^x\) are not the same. The first has a variable based with fixed power of \(2\). The other has a fixed base of \(2\) and a variable exponent.
Since equations with variable exponents are not the same as ones with fixed powers, we cannot use roots to solve them.
Complete methods for solving these exponential equations are covered in a precalculus course and involve something called logarithms. For now, however, we will rely on our background knowledge and employ a guess-and-check method.
(a) Solve \(2^x = 8\)
(b) Solve \(2^x = 32\)
(c) Solve \(3^x = 81\)
(d) Solve \(5^x = 125\)
Solution
Since the variables are in the exponent, these are all exponential equation. We will use guessing and checking to find the solutions.
(a) \(x = 3\) because \(2^3 = 8\).
(b) \(x = 5\) because \(2^5 = 32\).
(c) \(x = 4\) because \(3^4 = 81\).
(d) \(x = 3\) because \(5^3 = 125\).
Conclusion
The main point of this section is to emphasize the importance of recognizing the type of equation you are asked to solve. Knowing the type of equation helps you choose the proper tools for solving it.