Linear equations in one variable

Maintaining equation equality

Given any equation, if you multiply, divide, add, or subtract the same number on both sides, the equality of your new equation will hold. You can also raise both sides to the same power and keep equality. Lets look at an obviously true equation:

$$10=10$$

When we add 5 to each side the equality holds:

$$15=15$$

Now we can multiply each side by 4 to get:

$$60=60$$

If we divide both sides by 3 we have:

$$20=20$$

Subtracting 7 from both sides gives us:

$$13=13$$

As long as we do the same thing on both sides the equality of our equation will still be true. If we add a variable to the equation the rule holds. Observe:

$$\begin{align*} x & =2 \\ 3x & =6 \\ 3x+1 & =7 \\ \end{align*}$$

We multiplied both sides by 3 then added 1 to both sides, and you can easily substitute 2 in for x to check that the equality holds true. You could also start with the last equation and reverse those steps to reach the first equation, which is what we’ll be doing in this lesson.

Solving linear equations in one variable

Solving a linear equation in one variable means finding the value of the variable; this involves performing the same operations to both sides of an equation to maintain equality while working to isolate the variable on one side of the equation. In this example we solve for the variable x:

$$\begin{align*} 2(x-1) & = 13-x & \text{distribute the 2} \\ 2x-2 & = 13-x & \text{add x to both sides} \\ 3x-2 & = 13 & \text{add 2 to both sides} \\ 3x & = 15 & \text{divide both sides by 3} \\ x & = 5 \end{align*}$$

Being careful to perform each operation correctly is important, but the key to solving linear equations is figuring out which operation to perform next. Here’s a list of the usual order you’ll follow:

1. Distribute if necessary
2. Combine like terms if necessary
3. Add or subtract to move all x terms to one side
4. Add or subtract to move constants to the side opposite x
5. Divide both sides by the coefficient of x

Often times some of the steps will not be necessary. In the above example we used Step 1, then Step 3, then Step 4, then Step 5. Here’s an example Using only steps 4 and 5.

$$\begin{align*} & 4x+3=11 \\ & 4x=8 \\ & x=2 \end{align*}$$

When fractions are thrown into the mix the rules are the same but there are a couple things to note. In step 5 dividing by a fraction is computed by multiplying by the inverse of the fraction (remember the fraction rules? to divide by \( )\frac{2}{3} \) we multiply by \( \frac{3}{2}) \). Also, because of the difficulty of finding a common denominator it will sometimes save time to perform step 5 before step 4. This is optional and the following example will demonstrate:

$$\begin{align*} & \frac{1}{3}x+\frac{1}{6}=\frac{3}{10} \quad \text{step 4} \\ & \frac{1}{3}x=\frac{3}{10}-\frac{1}{6} \quad \text{find common denominator subtract and simplify} \\ & \frac{1}{3}x=\frac{18}{60}-\frac{10}{60}=\frac{8}{60}=\frac{2}{15} \quad \text{step 5} \\ & x=\frac{6}{15}=\frac{2}{5} \end{align*}$$

Solving this problem becomes simpler if we reverse the order of steps 4 and 5.

$$\begin{align*} & \frac{1}{3}x+\frac{1}{6}=\frac{3}{10} \qquad \text{step 5} \\ & x+\frac{1}{2}=\frac{9}{10} \qquad \quad \text{step 4}\\ & x=\frac{9}{10}-\frac{1}{2}=\frac{2}{5} \end{align*}$$

Whichever way you go you should arrive at the same answer.

Practice questions

Solve for the missing variable

1. \( 5x+7=52 \)
2. \( 3x-4=x-8 \)
3. \( 4x+10=7x-17 \)
4. \( 3(8-h)=5h+16 \)
5. \( 10x-7=5(x+1) \)
6. \( 2x-\frac{10}{3}=\frac{5}{6} \)
7. \( \frac{7}{2}x+\frac{5}{3}=\frac{5}{6} \)
8. \( \frac{1}{2}x+\frac{1}{3}=\frac{2}{3}x-\frac{1}{2} \)
9. A lawyer charges $200/hour for the first 10 hours and $100/hour for each additional hour. If a client’s bill comes to $2,250, how many hours did the lawyer work for that client?
10. A ship travels at 12 knots for x hours and at 16 knots for 21 hours. If the average speed of the ship during this voyage is 15 knots, what is x?

Solutions

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1. Solution: $$ \begin{align*} & 5x+7=52 \\ & 5x=45 \\ & x=9 \end{align*} $$
2. Solution: $$ \begin{align*} & 3x-4=x-8 \\ & 2x-4=-8 \\ & 2x=-4 \\ & x=-2 \end{align*} $$
3. Solution: $$ \begin{align*} & 4x+10=7x-17 \\ & 10=3x-17 \\ & 27=3x \\ & x=9 \end{align*} $$
4. Solution: $$ \begin{align*} & 3(8-h)=5h+16 \\ & 24-3h=5h+16 \\ & 24=8h+16 \\ & 8=8h \\ & h=1 \end{align*} $$
5. Solution: $$ \begin{align*} & 10x-7=5(x+1) \\ & 10x-7=5x+5 \\ & 5x-7=5 \\ & 5x=12 \\ & x=\frac{12}{5} \end{align*} $$
6. Solution: $$ \begin{align*} & 2x-\frac{10}{3}=\frac{5}{6} \\[0.5em] & 2x=\frac{5}{6}+\frac{20}{6} \\[0.5em] & 2x=\frac{25}{6} \\[0.5em] & x=\frac{25}{12} \end{align*} $$
7. Solution: $$ \begin{align*} & \frac{7}{2}x+\frac{5}{3}=\frac{5}{6} \\[0.5em] & \frac{7}{2}x=\frac{5}{6}-\frac{10}{6} \\[0.5em] & \frac{7}{2}x=-\frac{5}{6} \\[0.5em] & x=\frac{5}{6}\cdot\frac{2}{7} \\[0.5em] & x=-\frac{10}{42}=-\frac{5}{21} \end{align*} $$
8. Solution: $$ \begin{align*} & \frac{1}{2}x+\frac{1}{3}=\frac{2}{3}x-\frac{1}{2} \\[0.5em] & \frac{3}{6}x+\frac{2}{6}=\frac{4}{6}x-\frac{3}{6} \\[0.5em] & \frac{2}{6}=\frac{1}{6}x-\frac{3}{6} \\[0.5em] & \frac{5}{6}=\frac{1}{6}x \\[0.5em] & x=5 \end{align*} $$
9. Let \( x \) represent the number of hours billed to the client. Since the total bill is greater than \( $2,000 \) we know that \( x>10 \) so: $$ \begin{align*} & 200(10)+100(x-10)=2250 \\ & 2000+100x-1000=2250 \\ & 100x=1250 \\ & x=12.5 \text{hours} \end{align*} $$
10. We can write the average speed of the voyage both as \( 15 \) and as \( \frac{12x+16(21)}{x+21} \). Equating these gives us an equation we can solve for \( x \): $$ \begin{align*} & \frac{12x+16(21)}{x+21}=15 \\ & 12x+16(21)=15(x+21) \\ & 12x+16(21)=15x+15(21) \\ & 21=3x \\ & x=7 \end{align*} $$

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