1.19: Solving Linear Equations, Decimals, Rationals

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In this chapter we look at certain types of linear equations, those including decimal coefficients or rational coefficients. The reason why we discuss these separately is because we can “get rid” of the decimal numbers or denominators in the equation by performing a simple trick.

Recall from page 22 how we multiply decimal numbers by powers of 10 (that is \(10,100,1000, \ldots)\).

Example 17.1

Examples of multiplying by powers of 10:

\(0.05 \times 100=5\)
\(2.23 \times 10=22.3\)
\(0.7 \times 100=70\)
\(0.2 \times 10=2\)

Let’s look at the following linear equation with decimal coefficients:

\[0.02 y+0.1 y=2.4\nonumber\]

Step \(1 .\) Look at all the decimal numbers in the given equation: 0.02,0.1 and 2.4

Step 2. Pick the number \((s)\) with the most decimal place \((s)\) and count how many: 0.02 has two decimal places.

Step 3. Multiply both sides of the equation by \(100(1\) and two zeros) because the most number of decimal places is two.

\(\begin{align*} \text{Multiply both sides by }100: \quad 100 \times(0.02 y+0.1 y) &= 100 \times(2.4) \\ \text{Distribute}: \quad \quad \quad \quad 100 \times 0.02 y+100 \times 0.1 y &= 100 \times 2.4 \\ \Longrightarrow 2 y+10 y & = 240 \end{align*}\)

Step 4. Proceed to solve the linear equation as usual.

\[2 y+10 y=240\nonumber\]

\[\Longrightarrow 12 y=240\nonumber\]

\[\Longrightarrow y=\frac{240}{12}\nonumber\]

\[\Longrightarrow y=\frac{12 \cdot 20}{12}\nonumber\]

\[\Longrightarrow y=20\nonumber\]

Example 17.2

Solve the given equation:

a) \(1.4=0.2 x+4\)

Step 1. Look at all the decimal numbers in the given equation: 1.4 and 0.2

Step 2. Pick the number(s) with the most decimal places and count how many: 1.4 and 0.2 both have one decimal place.\

Step 3. Multiply both sides of the equation by \(10(1\) and one zero) because the most number of decimal places is one.

\(\begin{align*} \text{Multiply both sides by }100: \quad 100 10 \times(1.4) &= 10 \times(0.2 x+4) \\ \text{Distribute}: \quad \quad \quad \quad 10 \times 1.4 &= 10 \times 0.2 x+10 \times 4 \\ 14 & = 2 x+40 \end{align*}\)

Step 4. Proceed to solve the linear equation as usual.

\[14=2 x+40\nonumber\]

\[-40 \quad-40\nonumber\]

\[\Longrightarrow-26=2 x\nonumber\]

\[\Longrightarrow \frac{-26}{2}=\frac{2 x}{2}\nonumber\]

\[\Longrightarrow-13=x\nonumber\]

b) \(0.7+0.28 x=1.26\)

Step 1. Look at all the decimal numbers in the given equation: 0.7,0.28 and 1.26

Step 2. Pick the number(s) with the most decimal places and count how many: 0.28 and 1.26 both have two decimal places.

Step 3. Multiply both sides of the equation by \(100(1\) and two zeros) because the most number of decimal places is two.

\(\begin{align*} \text{Multiply both sides by }100: \quad 100 \times(0.7+0.28 x) &= 100 \times(1.26) \\ \text{Distribute}: \quad \quad \quad \quad 100 \times 0.7+100 \times 0.28 x &= 100 \times 1.26 \\ 70+28 x & = 126 \end{align*}\)

Step 4. Proceed to solve the linear equation as usual.

\[70+28 x=126\nonumber\]

\[-70 \quad-70\nonumber\]

\[\Longrightarrow 28 x=56\nonumber\]

\[\Longrightarrow \frac{28 x}{28}=\frac{56}{28}\nonumber\]

\[\Longrightarrow x=2\nonumber\]

c) \(0.5 x-0.235=0.06\)

Step 1. Look at all the decimal numbers in the given equation: 0.5, 0.235 and 0.06.

Step 2. Pick the number(s) with the most decimal places and count how many: 0.235 has three decimal places.

Step 3. Multiply both sides of the equation by 1000 (1 and three zeros) because the most number of decimal places is three.

\(\begin{align*} \text{Multiply both sides by }1000: \quad 1000 \times(0.5 x-0.235) &= 1000 \times(0.06) \\ \text{Distribute}: \quad \quad \quad \quad 1000 \times 0.5 x-1000 \times 0.235 &= 1000 \times 0.06 \\ 500 x-235 & = 60 \end{align*}\)

Step 4. Proceed to solve the linear equation as usual

\[500 x-235=60\nonumber\]

\[+235 \quad+235\nonumber\]

\[\Longrightarrow 500 x=295\nonumber\]

\[\Longrightarrow \frac{500 x}{500}=\frac{295}{500}\nonumber\]

\[\Longrightarrow x=\frac{5 \cdot 59}{5 \cdot 100}\nonumber\]

\[\Longrightarrow x=\frac{59}{100}\nonumber\]

In the last example, we could have proceeded as follows. Writing the equation using fractions gives

\[\frac{5}{10} x-\frac{235}{1000}=\frac{6}{100}\nonumber\]

Note that in the above example we multiplied by 1000 which is the least common denominator (the denominators being \(10,1000,\) and 100 ). We get

\[\frac{1000 \cdot 5}{10} x-\frac{1000 \cdot 235}{1000}=\frac{1000 \cdot 6}{100}\nonumber\]

Simplifying gives

\[100 \cdot 5 x-235=10 \cdot 6, \text { or equivalently, } 500 x-235=60\nonumber\]

This brings us to Step 4 above.

We can use this method to solve a linear equation involving fractions. Lets look at a couple of examples.

Example 17.3

Solving linear equations with rational coefficients

a) Solve \(\frac{1}{2}-\frac{3}{5} x=\frac{1}{6}\).

We could just treat this as in the last section but the arithmetic involves fractions. Subtract \(\frac{1}{2}\) from both sides (noting that \(\frac{1}{6}-\frac{1}{2}=\frac{1}{6}-\frac{3}{6}=-\frac{2}{6}=\) \(\left.-\frac{1}{3}\right)\)

\[-\frac{3}{5} x=-\frac{1}{3}\nonumber\]

Now multiplying both sides by \(-\frac{5}{3}\) gives

\[x=\left(-\frac{5}{3}\right)\left(-\frac{1}{3}\right)=\frac{5}{9}\nonumber\]

But we could also proceed by clearing fractions:

Step 1. List the denominators. The denominators are 2, 5, and 6.

Step 2. Find the least common denominator. The least common denominator is 30.

Step 3. Multiply both sides of the equation by the LCD (30) and simplify:

\[\frac{30 \cdot 1}{2}-\frac{30 \cdot 3}{5} x=\frac{30 \cdot 1}{6} \Longrightarrow 15-18 x=5\nonumber\]

Step 4. Solve as usual. We subtract 15 from both sides:

\[-18 x=-10\nonumber\]

Dividing by -18 gives

\[x=\frac{-10}{-18}=\frac{5}{9}\nonumber\]

b) Solve \(2-\frac{5 x}{3}=\frac{3 x}{5}-\frac{1}{6}\)

Step 1. Identify the denominators 3, 5, and 6.

Step 2. Find the least common denominator 30.

Step 3. Multiply both sides of the equation by 30 and simplify.

\[30 \cdot 2-\frac{30 \cdot 5 x}{3}=\frac{30 \cdot 3 x}{5}-\frac{30 \cdot 1}{6} \Longrightarrow 60-50 x=18 x-5\nonumber\]

Step 4. Solve as usual. Adding \(50 x\) to both sides gives

\[60=68 x-5\nonumber\]

Adding 5 to both sides gives

\[65=68 x\nonumber\]

Dividing by 68 gives

\[x=\frac{65}{68}\nonumber\]

Exit Problem

Solve: \(0.03 x+2.5 =4.27\)