Subsection 2.A.1 Solution to an Equation
In algebra, we are often presented with an equation whose solution is unknown. Finding the solution is called solving the equation
Example 2.A.1. Example Linear Equation.
\begin{align*}
LHS \amp = RHS\\
4x+16\amp =-4
\end{align*}
Equations are of the form "left-hand side" equals "right-hand side" or LHS\(=\)RHS for short\(.\checkmark\)
Notice the equation in Example 2.A.1 above has a missing part, or unknown number, that is marked by \(x.\) To show \(-5\) is a solution of this equation, we replace \(x\) by \(-5\) and simplify the LHS.
Example 2.A.2. A Solution.
Start with the equation \(4x + 16= -4.\)
Let \(x=-5\) in the LHS:
\begin{align*}
4x + 16\amp = -4\amp\amp\text{Substitute \(-5\) for \(x\) in the LHS}\\
4(-5) + 16\amp\amp\amp\text{Multiply \(4(-5)\)}\\
-20+ 16 \amp\amp\amp \text{Add \(-20+ 16\)}\\
-4 \amp\amp\amp \text{Compare the LHS to the RHS}\\
-4\amp = -4\amp\amp \text{LHS=RHS}
\end{align*}
Since we get a true statement \(-4=-4\text{,}\) \(x=-5\) IS a solution of the equation \(4x + 16= -4.\)\(\checkmark\)
Now the equation is a true statement! Notice also that if another number, for example, \(3\text{,}\) were plugged in for \(x\text{,}\) we would not get a true statement as seen in Example 2.A.3.
Example 2.A.3. NOT a Solution.
Start with the equation \(4x + 16= -4.\)
Let \(x=3\) in the LHS:
\begin{align*}
4x + 16\amp =-4\amp\amp\text{Substitute \(3\) for \(x\) in the LHS}\\
4(3) + 16 \amp\amp\amp \text{Multiply
\(4(3)\)}\\
12+ 16 \amp\amp\amp \text{Add
\(12+ 16\)}\\
28 \amp\amp\amp\text{Compare the LHS to the RHS}\\
28\amp \neq -4 \amp\amp \text{LHS}\neq\text{RHS}
\end{align*}
Since \(28\neq -4\text{,}\) \(x=3\) is NOT a solution of the equation \(4x + 16= -4.\)\(\checkmark\)
Due to the fact that \(LHS\neq RHS\text{,}\) this demonstrates that \(3\) is not the solution. However, depending on the complexity of the problem, this "guess and check" method is not very efficient. Thus, we take a more algebraic approach to solving equations. First we will focus on what are called "one-step equations" or equations that only require one step to solve. While these equations often seem very fundamental, it is important to master the pattern for solving these problems so we can solve more complex problems.
Subsection 2.A.2 Linear Equations - One-Step Equations
To solve equations, we perform mathematical operations to both sides of the equation to isolate the variable. For example, consider the following:
Example 2.A.4. Solve an Equation Involving Addition.
Solve the equation for \(x.\)
\begin{align*}
x+7 \amp = -5 \amp\amp \text{Notice the
\(7\) is added to the
\(x\)}\\
(x+7)-7 \amp = (-5) - 7 \amp\amp \text{Subtract
\(7\) from both sides to isolate
\(x\)}\\
x \amp = -12 \amp\amp \text{Our Solution}\checkmark
\end{align*}
WeBWorK: Entering Solutions to Equations.
Enter the entire expression x = -12, NOT just the value -12.
When you are solving an equation on paper, your final answer may be in the form \(3=t\text{.}\) In WeBWorK, you must enter your answer with the variable on the left side of the equal sign: t = 3
The terms in an equation may be in any order and the variable may be a letter other than \(x.\) The key is to focus on using operations that get the variable by itself on one side of the equals sign.
Example 2.A.5. Another Example With Addition.
Solve the equation for \(t.\)
In this example, \(5\) must be subtracted from both sides to solve for \(t.\)
\begin{align*}
8=5\amp +t\\
\underline{-5\,\,-5}\amp\\
3=t\amp
\end{align*}
Then we get our solution, \(t = 3.\checkmark\)
In a subtraction problem, we "counteract" negative numbers by adding them to both sides of the equation. For example, consider the following:
Example 2.A.6. Solve an Equation Involving Subtraction.
Solve the equation for \(x.\)
\begin{align*}
x-5\amp = 4 \amp\amp \text{Notice
\(5\) is subtracted from
\(x\)}\\
(x-5)+5\amp = 4+5 \amp\amp \text{Add
\(5\) to both sides of the equation to solve for
\(x\)}\\
x\amp = 9 \amp\amp \text{Our Solution}\checkmark
\end{align*}
With a multiplication problem, we "eliminate" a constant multiple by multiplying by the reciprocal (or dividing) on both sides. For example consider the following:
Example 2.A.7. Positive Constant Multiple.
Solve the equation for \(w.\)
\begin{align*}
4w \amp = 20 \amp\amp \text{Notice the constant multiple
\(4\) times the variable
\(w\)}\\
\frac{1}{4}(4w) \amp = \frac{1}{4}\cdot 20 \amp\amp \text{Multiply both sides by
\(\frac{1}{4}\)(or divide both sides by
\(4\))}\\
w \amp = 5 \amp\amp \text{Our Solution}\checkmark
\end{align*}
With multiplication problems it is very important that care is taken with signs. If the variable is multiplied by a negative number we multiply by its negative reciprocal (or divide by a negative.)
Example 2.A.8. Negative Constant Multiple.
Solve the equation for \(x.\)
\begin{align*}
-5x \amp = 30 \amp\amp \text{
\(x\) is multiplied by
\(-5\)}\\
-\frac{1}{5}(-5x) \amp = -\frac{1}{5}\cdot30 \amp\amp \text{Multiply both sides by
\(-\frac{1}{5}\)}\\
\amp\amp\amp\text{(or divide both sides by
\(-5\))}\\
x\amp =-6 \amp\amp \text{Our Solution}\checkmark
\end{align*}
The same process is used in each of the following examples. Notice how negative and positive numbers are handled as each problem is solved.
Example 2.A.9. Multiply by Reciprocals.
Solve each equation.
\begin{align*}
8t\amp =-24\\
\frac{1}{8}(8t)\amp =(-24)\frac{1}{8}\\
t\amp =-3
\end{align*}
\begin{align*}
-4z\amp =-20\\
\left(-\frac{1}{4}\right)(-4z)\amp =(-20)\left(-\frac{1}{4}\right)\\
z\amp =5
\end{align*}
\begin{align*}
42\amp =7x\\
\frac{1}{7}(42)\amp =\frac{1}{7}(7x)\\
6\amp =x
\end{align*}
Checkpoint 2.A.10. Solve One-Step Equation.
In division problems, "eliminate the denominator" by multiplying on both sides. For example consider:
Example 2.A.11. Eliminate Denominator.
Solve the equation for \(t.\)
\begin{align*}
\frac{t}{5}\amp =-3 \amp\amp \text{Variable
\(t\) is divided by
\(5\)}\\
(5)\frac{t}{5}\amp =-3(5) \amp\amp \text{Multiply both sides by
\(5\)}\\
t\amp =-15 \amp\amp \text{Our Solution}\checkmark
\end{align*}
The same process is used in each of the following examples.
Example 2.A.12. Eliminate Denominators.
Solve for each variable.
\begin{align*}
\frac{x}{-7}\amp =-2\\
(-7)\frac{x}{-7}\amp =-2(-7)\\
x\amp=14
\end{align*}
\begin{align*}
\frac{t}{8}\amp =5\\
(8)\frac{t}{8}\amp =5(8)\\
t\amp =40
\end{align*}
\begin{align*}
\frac{w}{-4}\amp =9\\
(-4)\frac{w}{-4}\amp =9(-4)\\
w\amp =-36
\end{align*}
Checkpoint 2.A.13. Solve One-Step Equation.
The process described above is fundamental to solving equations. Soon, the problems we see will have several more steps. These problems may seem more complex, but the process and patterns used will remain the same.
Subsection 2.A.3 Linear Equations - Two-Step Equations
After mastering the technique for solving equations that are one-step equations, we are ready to consider two-step equations. As we solve two-step equations, the important thing to remember is that everything works backwards! When working with one-step equations, we learned that in order to clear a "plus five" in the equation, we would subtract five from both sides. We learned that to clear "divided by seven" we multiply by seven on both sides. The same pattern applies to the order of operations. When solving for our variable, we use order of operations backwards as well. This means we will add or subtract first, then multiply or divide second (then exponents, and finally any parentheses or grouping symbols, but that's another lesson).
Example 2.A.14. Addition then Division.
Solve the equation for \(.\)
\begin{gather*}
4x-20=-8
\end{gather*}
We have two numbers on the same side as the variable \(x.\) We need to handle the \(4\) and the \(-20.\) We know to "eliminate" the four we need to divide, and to "eliminate" the subtracted twenty we will add twenty to both sides. If order of operations is done backwards, we add or subtract first. Therefore we will add \(20\) to both sides first. Once we are done with that, we will divide both sides by \(4.\) The steps are shown below.
\begin{align*}
4x\color{red}{-20}\color{black}{}\amp= -8 \amp\amp \text{Start by focusing on the subtract
\(20\)}\\
\underline{+20}\amp\phantom{12}\underline{+20} \amp\amp \text{Add
\(20\) to both sides}\\
4x\amp = 12 \amp\amp \text{Now we focus on the
\(4\) multiplied by
\(x\)}\\
\overline{\,4\,}\amp\phantom{123}\overline{\,4\,} \amp\amp \text{Divide both sides by
\(4\)}\\
x\amp =3 \amp\amp \text{Our Solution}\checkmark
\end{align*}
To check the correctness of Our Solution we substitute \(x=3\) back in to the original equation:
\begin{align*}
4x-20 \amp = -8 \amp\amp \text{Substitute
\(x=3\) into the LHS}\\
4(3)-20 \amp\amp\amp \text{Multiply
\(4(3)\)}\\
12-20 \amp\amp\amp \text{Subtract
\(12-20\)}\\
-8 \amp\amp\amp\text{Compare the LHS to the RHS}\\
-8\amp =-8\amp\amp\text{LHS=RHS}
\end{align*}
Since we get a true statement \(-8=-8\text{,}\) Our Solution \(x=3\) is correct.\(\checkmark\)
The same process is used to solve any two-step equations. Add or subtract first, then multiply or divide. Consider our next example and notice how the same process is applied.
Example 2.A.15. Subtraction then Division.
Solve the equation for \(P.\)
\begin{align*}
5P\color{red}{+7}\color{black}{}\amp =7 \amp\amp \text{Start by focusing on the plus
\(7\)}\\
\underline{-7}\amp\phantom{1}\underline{-7} \amp\amp \text{Subtract
\(7\) from both sides}\\
5P\amp = 0 \amp\amp \text{Now we focus on the
\(5\) multiplied by
\(P\)}\\
\overline{\,5\,}\phantom{1}\amp\phantom{12} \overline{\,5\,} \amp\amp \text{Divide both sides by
\(5\) (recall
\(0/5=0\))}\\
P\amp= 0 \amp\amp \text{Our Solution}\checkmark
\end{align*}
Notice the seven subtracted out completely! Many students get stuck on this point, do not forget that we have a number for "nothing left" and that number is zero.
When solving two-step equations, remember the sign always stays with the number. Consider the following example.
Example 2.A.16. Negative Coefficient.
\begin{align*}
\color{red}{4}\color{black}{-2t}\amp = 10 \amp\amp \text{Start by focusing on the positive
\(4\)}\\
\underline{-4\phantom{1234}}\amp\underline{\phantom{1}-4} \amp\amp \text{Subtract
\(4\) from both sides}\\
-2t\amp =6 \amp\amp\text{Notice
\(-2\) multiplies
\(t\) (a negative coefficient)}\\
\overline{-2}\amp\phantom{12}\overline{-2} \amp\amp\text{Divide both sides by
\(-2\)}\\
t\amp =-3 \amp\amp \text{Our Solution}\checkmark
\end{align*}
The same is true even if the coefficient in front of the variable is a negative \(1\). Consider the next example.
Example 2.A.17. Negative Coefficient.
\begin{align*}
\color{red}{8}\color{black}{-x}\amp =2 \amp\amp\text{Start by focusing on the positive
\(8\)}\\
\underline{-8\phantom{123}}\amp\underline{\phantom{1}-8}\amp\amp\text{Subtract
\(8\) from both sides}\\
-x\amp = -6\amp\amp \text{Notice \(-x\) is
\(-1\) times
\(x\)}\\
(-1)(-x)\amp = -6(-1) \amp\amp \text{Multiply both sides by
\(-1\)}\\
\amp\amp\amp\text{(or divide both sides by
\(-1\)) for the same result}\\
x\amp =6 \amp\amp\text{Our Solution}\checkmark
\end{align*}
The first step is to add or subtract, the second is to multiply or divide. This pattern is seen in each of the following three examples.
Example 2.A.18. More Two-Step Examples.
Solve each equation.
\begin{align*}
7-5y=17\amp\\
\underline{-7\phantom{12345}-7}\amp\\
-5y=10\amp\\
\overline{-5}\phantom{123}\overline{-5}\amp\\
y=-2\amp
\end{align*}
\begin{align*}
8=10+2t\amp\\
\underline{-10\phantom{1}-10}\phantom{1234}\amp\\
-2=2t\phantom{1234}\amp\\
\overline{2}\phantom{1234}\overline{2}\phantom{1234}\amp\\
-1=t\phantom{12345}\amp
\end{align*}
\begin{align*}
-3=\frac{x}{5}-4\amp\\
\underline{+4\phantom{12345}+4}\amp\\
1=\frac{x}{5}\phantom{123}\amp\\
(5)1=\frac{x}{5}(5)\amp\\
5=x\phantom{123}\amp
\end{align*}
Checkpoint 2.A.19. Solve Two-Step Equation.
Subsection 2.A.4 Solving Linear Equations-General Equations
Often as we are solving linear equations we will need to do some work to set them up into a form we are familiar with solving. This section will focus on manipulating an equation so that we can use our pattern for solving two-step equations.
To handle the parentheses we use the distributive property. This is shown in the following example.
Example 2.A.20. Distribute First.
Solve the equation for \(x.\)
\begin{align*}
4(2x-6)\amp =16 \amp\amp \text{Distribute
\(4\) through parentheses: multiply
\(4(2)\) and
\(4(-6)\)}\\
8x-24\amp =16 \amp\amp \text{Focus on the subtraction first}\\
\underline{+24}\amp\underline{\phantom{1}+ 24} \amp\amp \text{Add
\(24\) to both sides}\\
8x\amp = 40 \amp\amp \text{Now focus on the multiplication by
\(8\)}\\
\overline{\,8\,}\amp\phantom{123}\overline{\,8\,} \amp\amp \text{Divide both sides by
\(8\)}\\
x\amp =5 \amp\amp \text{Our Solution}\checkmark
\end{align*}
Often after we distribute there will be some like terms on one side of the equation. Example 2.A.21 shows distributing to "clear" the parentheses followed by combining like terms. Notice we only combine like terms on the same side of the equation. Once we have done this, our next example solves just like any other two-step equation.
Example 2.A.21. Distribute First.
Solve the equation for \(t.\)
\begin{align*}
3(2t-4)+9\amp =15 \amp\amp \text{Distribute the
\(3\) through the parentheses}\\
6t-12+9\amp =15 \amp\amp \text{Combine like terms,
\(-12+9\)}\\
6t-3\amp =15 \amp\amp \text{Focus on the subtraction first}\\
\underline{+3}\amp\underline{\phantom{1}+3} \amp\amp \text{Add
\(3\) to both sides}\\
6t\amp = 18 \amp\amp \text{Now focus on multiplication by
\(6\)}\\
\overline{\,6\,}\amp\phantom{123}\overline{\,6\,} \amp\amp \text{Divide both sides by
\(6\)}\\
t\amp = 3 \amp\amp \text{Our Solution}\checkmark
\end{align*}
We may see the variable on both sides of an equation.
Example 2.A.22. Collect Like Terms First.
Solve the equation for \(x.\)
\begin{gather*}
4x-6=2x+10
\end{gather*}
Notice here the \(x\) is on both the left and right sides of the equation. This can make it difficult to decide which side to work with. We fix this by adding or subtracting one of the terms with \(x\) to the other side, much like we do with a constant term. It doesn't matter which term gets added or subtracted, \(4x\) or \(2x\text{,}\) however, it would be our suggestion to subtract the smaller term in this case (to avoid negative coefficients). For this reason we begin this problem by subtracting \(2x\) from both sides.
\begin{align*}
4x-6\amp =2x+10 \amp\amp \text{Notice the variable
\(x\) on both sides}\\
\underline{-2x\phantom{123}}\amp\underline{\phantom{1}-2x} \amp\amp \text{Subtract
\(2x\) from both sides}\\
2x-6\amp =10 \amp\amp \text{Focus on the subtraction first}\\
\underline{+6}\amp\underline{\phantom{1}+6} \amp\amp \text{Add
\(6\) to both sides}\\
2x\amp =16 \amp\amp \text{Focus on the multiplication by
\(2\)}\\
\overline{\,2\,}\amp\phantom{123}\overline{\,2\,} \amp\amp \text{Divide both sides by
\(2\)}\\
x\amp =8 \amp\amp \text{Our Solution}\checkmark
\end{align*}
The next example shows the check on the solution in Example 2.A.22 . Here Our Solution \(x=8\) is substituted for the \(x\) on both the LHS and RHS before simplifying.
Example 2.A.23. Check Our Solution \(x=8\).
Check that \(x=8\) is a solution of the equation \(4x-6=2x+10.\)
\begin{align*}
\text{LHS: }\amp\amp\amp 4x-6\\
4(8)-6 \amp\amp\amp \text{Multiply
\(4(8)\)}\\
32 -6 \amp\amp\amp \text{Subtract}\\
26 \amp\amp\amp \text{LHS value}
\end{align*}
\begin{align*}
\text{RHS: }\amp\amp\amp 2x+10\\
2(8)+10 \amp\amp\amp \text{Multiply
\(2(8)\)}\\
16+10 \amp\amp\amp \text{Add}\\
26 \amp\amp\amp \text{RHS value}
\end{align*}
Since \(26=26\) we have LHS=RHS which means Our Solution \(x=8\) is correct!\(\checkmark\)
The next example illustrates the same process with negative coefficients. Notice first the negative term with the variable is added to both sides.
Example 2.A.24. Collect Like Terms.
Solve the equation for \(z.\)
\begin{align*}
-3z+9\amp =6z-27 \amp\amp \text{Notice the variable
\(z\) on both sides,
\(-3z\) is negative}\\
\underline{+3z\phantom{123}}\amp\phantom{1}\underline{+3z} \amp\amp \text{Add
\(3z\) to both sides}\\
9\amp =9z-27 \amp\amp \text{Focus on the subtraction of
\(27\) first}\\
\underline{+27}\amp\underline{\phantom{12345} +27} \amp\amp \text{Add
\(27\) to both sides}\\
36\amp =9z \amp\amp \text{Focus on the multiplication by
\(9\)}\\
\overline{\,9\,}\amp\phantom{123}\overline{\,9\,} \amp\amp \text{Divide both sides by
\(9\)}\\
4\amp =z \amp\amp \text{ }\\
z\amp =4 \amp\amp \text{Our Solution}\checkmark
\end{align*}
In the following problems we have parentheses and the variable on both sides. Notice in each of the following examples we distribute, then combine like terms, then move the variable to one side of the equation.
Example 2.A.25. Distribute and Collect.
Solve the equation for \(x.\)
\begin{align*}
2(x-5)+3x\amp =x+18 \amp\amp \text{Distribute the
\(2\) through parentheses}\\
2x-10+3x\amp =x+18 \amp\amp \text{Combine like terms
\(2x+ 3x\)}\\
5x-10\amp =x+18 \amp\amp \text{Notice the variable
\(x\) on both sides,
\(x\) is smaller}\\
\underline{-x\phantom{12345}}\amp\underline{\phantom{1}-x} \amp\amp \text{Subtract
\(x\) from both sides}\\
4x-10\amp =18 \amp\amp \text{Focus on the subtraction of
\(10\)}\\
\underline{+10}\amp\underline{\phantom{1}+10} \amp\amp \text{Add
\(10\) to both sides}\\
4x\amp =28 \amp\amp \text{Focus on the multiplication by
\(4\)}\\
\overline{\,4\,}\amp\phantom{123}\overline{\,4\,} \amp\amp \text{Divide both sides by
\(4\)}\\
x\amp =7 \amp\amp \text{Our Solution}\checkmark
\end{align*}
Checkpoint 2.A.26. Solve Equation.
Equations with fractions are often easier to solve if we eliminate the denominators first.
Example 2.A.27. Equation with Fractions.
Solve the equation for \(x.\)
\begin{align*}
\dfrac{x}{6}\amp =7-\dfrac{x}{2} \amp\amp \text{Note that \(6\) is the LCD for \(\frac{x}{6}\) and \(\frac{x}{2}\)}\\
6\left(\dfrac{x}{6}\right)\amp =6\left( 7-\dfrac{x}{2}\right) \amp\amp \text{If we multiply both sides by \(6\),}\\
\amp\amp\amp\text{both deniminators will be eliminated}\\
x\amp =6(7)-6\left(\dfrac{x}{2}\right) \amp\amp \text{Distribute the \(6\) and simplify}\\
x\amp =42-3x \amp\amp \text{Since \(6\left(\dfrac{x}{2}\right) =\frac{6}{2}x = 3x\)}\\
\underline{+3x}\amp\underline{\phantom{1234}+3x} \amp\amp \text{Add \(3x\) to both sides}\\
4x\amp =42 \amp\amp \text{Simplify}\\
\overline{\,4\,}\amp\phantom{123}\overline{\,4\,} \amp\amp \text{Divide both sides by \(4\)}\\
x\amp =\frac{42}{4} \amp\amp \text{Reduce the fraction}\\
x\amp =\frac{21}{2} \amp\amp \text{Our Solution}\checkmark
\end{align*}
Checkpoint 2.A.28. Solve Equation with Fractions.
Example 2.A.29. Equation with Fractions.
Solve the equation for \(x.\)
\begin{align*}
x -\dfrac{x+7}{8}\amp =\dfrac{5x}{12} \amp\amp \text{Note that \(24\) is the LCD for \(\frac{x+7}{8}\) and \(\frac{5x}{12}\)}\\
24\left(x -\dfrac{x+7}{8}\right)\amp = 24\left( \dfrac{5x}{12} \right) \amp\amp \text{Multiply both sides by \(24\),}\\
\amp\amp\amp\text{to eliminate both deniminators.}\\
24x -3(x+7)\amp = 10x \amp\amp \text{Distribute and simplify}\\
24x -3x-21\amp = 10x \amp\amp \text{ }\\
21x-21\amp = 10x \amp\amp \text{ }\\
\underline{-10x}\amp\underline{\phantom{12}-10x} \amp\amp \text{Subtract \(10x\) from both sides}\\
11x-21\amp = 0 \amp\amp \text{ }\\
\underline{+21}\amp\underline{\phantom{123}+21} \amp\amp \text{Add \(21\) to both sides}\\
11x\amp = -21 \amp\amp \text{Simplify}\\
\overline{\,11\,}\amp\phantom{123}\overline{\,11\,} \amp\amp \text{Divide both sides by \(11\)}\\
x\amp =\frac{-21}{11} \amp\amp \text{Our Solution}\checkmark
\end{align*}
Sometimes we may have to distribute more than once to clear several parentheses. Remember to combine like terms after you distribute!
Example 2.A.30. Distribute and Collect.
Solve the equation for \(y.\)
\begin{align*}
3(4y-5)-4(2y+1)\amp =5 \amp\amp \text{Distribute the
\(3\) and
\(-4\) through parentheses}\\
12y-15-8y-4\amp =5 \amp\amp \text{Combine like terms
\(12y-8y\) and
\(-15-4\)}\\
4y-19\amp =5 \amp\amp \text{Focus on the subtraction of
\(19\)}\\
\underline{+19}\amp\underline{\, +19} \amp\amp \text{Add
\(19\) to both sides}\\
4y\amp =24 \amp\amp \text{Focus on the multiplication by
\(4\)}\\
\overline{\,4\,}\amp\phantom{123}\overline{\,4\,} \amp\amp \text{Divide both sides by
\(4\)}\\
y\amp =6 \amp\amp \text{Our Solution}\checkmark
\end{align*}
This leads to a \(5\)-step process to solve any linear equation. While all five steps aren't always needed, this can serve as a guide to solving equations.
- Distribute through any parentheses.
- Combine like terms on each side of the equation.
- Get the variable on one side by adding or subtracting.
- Solve the remaining 2-step equation (add or subtract then multiply or divide).
- Check your answer by plugging it back in for \(x\) to find a true statement.
We can see each of the above five steps worked through our next example.
Example 2.A.31. Demonstrate Five-Step Solution.
Solve the equation for \(x.\)
\begin{align*}
4(2x-6)+9\amp =3(x-7)+8x \amp\amp \text{Step 1: Distribute
\(4\) and
\(3\) through parentheses}\\
8x-24+9\amp =3x-21+8x \amp\amp \text{Step 2: Combine like terms
\(-24+9\) and
\(3x+8x\)}\\
8x-15\amp =11x-21 \amp\amp \text{Notice the variable
\(x\) is on both sides}\\
\underline{-8x\phantom{1234}}\amp\underline{\phantom{1}-8x} \amp\amp \text{Step 3: Subtract
\(8x\) from both sides}\\
-15\amp =3x-21 \amp\amp \text{Step 4: Focus on the subtraction of
\(21\)}\\
\underline{+21}\amp\underline{\phantom{12345} +21} \amp\amp \text{Add
\(21\) to both sides}\\
6\amp =3x \amp\amp \text{Focus on the multiplication by
\(3\)}\\
\overline{\,3\,}\amp\phantom{123}\overline{\,3\,} \amp\amp \text{Divide both sides by
\(3\)}\\
2\amp =x \amp\amp \text{ }\\
x\amp =2 \amp\amp \text{Our Solution}\checkmark
\end{align*}
We leave Step 5 for you: check that \(x=2\) IS a solution by plugging this value back in to the original equation.
Let's try one involving scientific notation. The fundamental ideas of solving an equation still apply.
Example 2.A.32. Solve an Equation Involving Scientific Notation.
Solve the equation for \(x.\)
\begin{align*}
\left(7\times 10^{3}\right)x = 1.4\times 10^{-7}\amp\amp\amp \text{Divide both sides by } 7\times 10^{3}\\
x = \dfrac{1.4\times 10^{-7}}{7\times 10^{3}} \amp\amp\amp \text{Deal with numbers and
\(10\)'s separately}\\
\dfrac{1.4}{7}=0.2 \amp\amp\amp \text{Divide the numbers}\\
\dfrac{10^{-7}}{10^3}=10^{-7-3}=10^{-10} \amp\amp\amp \text{Simplify the powers on } 10\\
x = 0.2 \times 10^{-10} \amp\amp\amp \text{Note: The answer is not in correct scientific notation.}\\
x = 2.0 \times 10^{-11} \amp\amp\amp \text{Move the decimal one place to the right, }\\
\amp\amp\amp \text{decrease the power on \(10\) by \(1\).}\\
x=2.0 \times 10^{-11} \amp\amp\amp \text{Our Solution}\checkmark
\end{align*}
Checkpoint 2.A.33. Solve Equation with Scientific Notation.
Subsection 2.A.5 Absolute Value Equations
Absolute value is defined as distance from zero. Consider both integers \(5\) and \(-5.\) Since both positive and negative \(5\) have a distance of \(5\) units from \(0\) on the number line, both have an absolute value of \(5.\)
\begin{align*}
\lvert 5\rvert \amp =5 \text{ and}\\
\lvert -5\rvert \amp =5
\end{align*}
Checkpoint 2.A.34. Determine the Absolute Value.
Thus when solving equations with absolute value we can end up with more than one possible answer. This is because the expression inside the absolute value can be either positive or negative and we must account for both possibilities when solving equations. This is illustrated in the following example.
Example 2.A.35. Absolute Value Equation: Two Solutions.
\begin{align*}
\lvert x\rvert = 5 \amp\amp\amp \text{Absolute value can be positive or negative}\\
\text{
\(x=5\) or
\(x=-5\)} \amp\amp\amp \text{Our Solution}\checkmark
\end{align*}
Notice that we have determined two possibilities, both the positive and negative. Either way, the absolute value of our number will be positive \(5.\)
WeBWorK: Entering Two Solutions.
Be sure to include the word "or" between the solutions: x = 5 or x = -5.
Checkpoint 2.A.36. Solve the Absolute Value Equation.
Absolute value may also be used to define the distance from a point other than zero. Note that the "distance" between two numbers is the positive difference between the numbers—the absolute value.
Example 2.A.37. Distance Equation.
Suppose the distance between \(x\) and \(2\) is exactly \(3.\) Determine \(x\) and write an absolute value equation that describes this situation.
The difference between \(x\) and \(2\) is \(x-2\) and the difference between the numbers could be to the left (\(-3\)) or to the right (\(+3\)) of \(2\text{.}\)
\begin{align*}
\text{ Solve: } \amp x-2=-3 \amp\amp \text{ Solve: } \amp x-2=3\\
\amp x = -3+2 = -1 \amp\amp \amp x = 3+2 = 5\\
\amp \text{\(-1\) is \(3\) units to the left of \(2\)} \amp\amp \amp \text{ \(5\) is \(3\) units to the right of \(2\)}
\end{align*}
\begin{align*}
\amp x=-1 \amp \text{ or } \amp x=5 \amp \text{Our Solution}\checkmark
\end{align*}
To denote "the distance between the number \(x\) and \(2\) is \(3\)", we write the difference in absolute value.
\begin{align*}
\lvert x-2\rvert =3\amp\amp \text{Absolute value equation that describes this situation}\checkmark
\end{align*}
Checkpoint 2.A.38. Absolute Value Equations.
The next few examples demonstrate the steps to solve a general absolute value equation. Remember that the expression inside the absolute value symbols may be positive or negative—the absolute value will make the value non-negative in the end—so, both possibilities must be solved for to determine all solutions to an absolute value equation.
Example 2.A.39. Absolute Value of an Expression.
Solve the equation for \(x.\)
\begin{align*}
\lvert 2x-1\rvert = 9 \amp\amp\amp \text{Consider both positive or negative values}\\
\text{
\(2x-1=9\) or
\(2x-1=-9\)} \amp\amp\amp \text{Two equations to solve}
\end{align*}
Now notice we have two equations to solve, each equation will give us a different solution. Both equations solve (separately) like any other two-step equation.
\begin{align*}
2x-1=9\amp\amp\amp\text{ or }\amp\amp\amp 2x-1=-9\\
\underline{+1\,+1}\amp\amp\amp\text{ }\amp\amp\amp\underline{+1\phantom{12}+1}\\
2x=10\amp\amp\amp\text{ }\amp\amp\amp 2x=-8\\
\overline{2}\phantom{1234}\overline{2}\amp\amp\amp\phantom{1234}\amp\amp\amp\overline{2}\phantom{1234}\overline{\,2\,}\\
x=5\amp\amp\amp\text{ }\amp\amp\amp x=-4\\
\amp\amp x=5\amp\text{ or } x=-4\amp\amp\amp\text{Our Solution}\checkmark
\end{align*}
Checkpoint 2.A.40. Solve Absolute Value Equation.
Remember the result of an absolute value must always be nonnegative. Notice what happens in the next example.
Example 2.A.41. Absolute Values Can't Be Negative.
\begin{align*}
\lvert 2x-5\rvert = -4 \amp\amp\amp \text{Notice the problem asks where the expression is negative}\\
\amp\amp\amp \text{STOP! Result of absolute value can't be negative! }\\
\text{No solution} \amp\amp\amp \text{Our Solution}\checkmark
\end{align*}
Notice in the above example, the absolute value expression equals a negative number! This is impossible with absolute value. When this occurs we STOP and conclude there is no solution.
WeBWorK: Enter "No solution".
If the equation has no solution, type No solution.
There is one more special case to consider.
Example 2.A.42. Absolute Value Equals Zero.
\begin{align*}
\lvert x-3\rvert = 0 \amp\amp\amp \text{What numbers are zero distance from \(3\)?}\\
x = 3 \amp\amp\amp \text{Only \(3\) is zero distance from itself }\checkmark\\
\amp\amp\amp \\
\amp\amp\amp \text{Or we could solve this another way:}\\
x-3 = 0 \amp\amp\amp \text{Since \(-0=0\) we only need one equation to solve}\\
x=3 \amp\amp\amp \text{Add \(3\) to both sides}\checkmark
\end{align*}
Checkpoint 2.A.43. Absolute Value Special Cases.
