Multi-Step Linear Equations
2.2 Learning Objectives
- Use properties of equality to isolate variables and solve algebraic equations
- Solve equations containing absolute values
- Use the properties of equality and the distributive property to solve equations containing parentheses
- Clear fractions and decimals from equations to make them easier to solve
- Solve equations that have one solution, no solution, or an infinite number of solutions
- Recognize when a linear equation that contains absolute value does not have a solution
2.2.1 Use properties of equality to isolate variables and solve algebraic equations
There are some equations that you can solve in your head quickly, but other equations are more complicated. Multi-step equations, ones that takes several steps to solve, can still be simplified and solved by applying basic algebraic rules such as the multiplication and addition properties of equality. In this section we will explore methods for solving multi-step equations that contain grouping symbols and several mathematical operations. We will also learn techniques for solving multi-step equations that contain absolute values. Finally, we will learn that some equations have no solutions, while others have an infinite number of solutions. First, let's define some important terminology:- variables: variables are symbols that stand for an unknown quantity, they are often represented with letters, like x, y, or z.
- coefficient: Sometimes a variable is multiplied by a number. This number is called the coefficient of the variable. For example, the coefficient of 3x is 3.
- term: a single number, or variables and numbers connected by multiplication. -4, 6x and [latex]x^2[/latex] are all terms
- expression: groups of terms connected by addition and subtraction. [latex]2x^2-5[/latex] is an expression
- equation: an equation is a mathematical statement that two expressions are equal. An equation will always contain an equal sign with an expression on each side. Think of an equal sign as meaning "the same as." Some examples of equations are [latex]y = mx +b\\[/latex], [latex]\frac{3}{4}r = v^{3} - r\\[/latex], and [latex]2(6-d) + f(3 +k) = \frac{1}{4}d\\[/latex]
Example 2.2.A
Solve [latex]3y+2=11[/latex].Answer: Subtract 2 from both sides of the equation to get the term with the variable by itself.
[latex] \displaystyle \begin{array}{r}3y+2\,\,\,=\,\,11\\\underline{\,\,\,\,\,\,\,-2\,\,\,\,\,\,\,\,-2}\\3y\,\,\,\,=\,\,\,\,\,9\end{array}[/latex]
Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.[latex]\begin{array}{r}\,\,\,\,\,\,\underline{3y}\,\,\,\,=\,\,\,\,\,\underline{9}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\\\,\,\,\,\,\,\,\,\,\,y\,\,\,\,=\,\,\,\,3\end{array}[/latex]
Answer
[latex]y=3[/latex]Example 2.2.B
Solve [latex]3x+5x+4-x+7=88[/latex].Answer: There are three like terms [latex]3x[/latex], [latex]5x[/latex], and [latex]–x[/latex] involving a variable. Combine these like terms. 4 and 7 are also like terms and can be added.
[latex]\begin{array}{r}\,\,3x+5x+4-x+7=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x+4+7=\,\,\,88\end{array}[/latex]
The equation is now in the form [latex]ax+b=c[/latex], so we can solve as before.[latex]7x+11\,\,\,=\,\,\,88[/latex]
Subtract 11 from both sides.[latex]\begin{array}{r}7x+11\,\,\,=\,\,\,88\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-11\,\,\,\,\,\,\,-11}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x\,\,\,=\,\,\,77\end{array}[/latex]
Divide both sides by 7.[latex]\begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{7x}\,\,\,=\,\,\,\underline{77}\\7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,=\,\,\,11\end{array}[/latex]
Answer
[latex]x=11[/latex]Example 2.2.C
Solve: [latex]4x-6=2x+10[/latex]Answer: Choose the variable term to move—to avoid negative terms choose [latex]2x[/latex]
[latex]\,\,\,4x-6=2x+10\\\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x}\\\,\,\,2x-6=10[/latex]
Now add 6 to both sides to isolate the term with the variable.
[latex]\begin{array}{r}2x-6=10\\\underline{\,\,\,\,+6\,\,\,+6}\\2x=16\end{array}[/latex]
Now divide each side by 2 to isolate the variable x.
[latex]\begin{array}{c}\frac{2x}{2}=\frac{16}{2}\\\\x=8\end{array}[/latex]
2.2.2 Solving Multi-Step Equations With Absolute Value
We can apply the same techniques we used for solving a one-step equation which contained absolute value to an equation that will take more than one step to solve. Let's start with an example where the first step is to write two equations, one equal to positive 26 and one equal to negative 26.Example 2.2.D
Solve for p. [latex]\left|2p–4\right|=26[/latex]Answer: Write the two equations that will give an absolute value of 26.
[latex] \displaystyle 2p-4=26\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,2p-4=\,-26[/latex]
Solve each equation for p by isolating the variable.[latex] \displaystyle \begin{array}{r}2p-4=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2p-4=\,-26\\\underline{\,\,\,\,\,\,+4\,\,\,\,+4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,+4\,\,\,\,\,\,\,+4}\\\underline{2p}\,\,\,\,\,\,=\underline{30}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{2p}\,\,\,\,\,=\,\underline{-22}\\2\,\,\,\,\,\,\,=\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,2\\\,\,\,\,\,\,\,\,\,p=15\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=\,-11\end{array}[/latex]
Check the solutions in the original equation.[latex] \displaystyle \begin{array}{r}\,\,\,\,\,\left| 2p-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 2p-4 \right|=26\\\left| 2(15)-4 \right|=26\,\,\,\,\,\,\,\left| 2(-11)-4 \right|=26\\\,\,\,\,\,\left| 30-4 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -22-4 \right|=26\\\,\,\,\,\,\,\,\,\,\,\,\,\left| 26 \right|=26\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| -26 \right|=26\end{array}[/latex]
Both solutions check!Answer
[latex]p=15[/latex] or [latex]p=-11[/latex]Example 2.2.E
Solve for w. [latex]3\left|4w–1\right|–5=10[/latex]Answer: Isolate the term with the absolute value by adding 5 to both sides.
[latex]\begin{array}{r}3\left|4w-1\right|-5=10\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,+5\,\,\,+5}\\ 3\left|4w-1\right|=15\end{array}[/latex]
Divide both sides by 3. Now the absolute value is isolated.[latex]\begin{array}{r} \underline{3\left|4w-1\right|}=\underline{15}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\\\left|4w-1\right|=\,\,5\end{array}[/latex]
Write the two equations that will give an absolute value of 5 and solve them.[latex] \displaystyle \begin{array}{r}4w-1=5\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,4w-1=-5\\\underline{\,\,\,\,\,\,\,+1\,\,+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,\,\,+1\,\,\,\,\,+1}\\\,\,\,\,\,\underline{4w}=\underline{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{4w}\,\,\,\,\,\,\,=\underline{-4}\\4\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\\\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\text{or}\,\,\,\,\,-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Check the solutions in the original equation.[latex] \displaystyle \begin{array}{r}\,\,\,\,\,3\left| 4w-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\3\left| 4\left( \frac{3}{2} \right)-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| 4w-1\, \right|-5=10\\\\\,\,\,\,\,\,3\left| \frac{12}{2}-1\, \right|-5=10\,\,\,\,\,\,\,3\left| 4(-1)-1\, \right|-5=10\\\\\,\,\,\,\,\,\,\,3\left| 6-1\, \right|-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -4-1\, \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left(5\right)-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left| -5 \right|-5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15-5=10\\10=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10=10\end{array}[/latex]
Both solutions check!Answer
[latex]w=-1\,\,\,\,\text{or}\,\,\,\,w=\frac{3}{2}[/latex]2.2.3 The Distributive Property
As we solve linear equations, we often need to do some work to write the linear equations in a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution. Parentheses can make solving a problem difficult, if not impossible. To get rid of these unwanted parentheses we have the distributive property. Using this property we multiply the number in front of the parentheses by each term inside of the parentheses.The Distributive Property of Multiplication
For all real numbers a, b, and c, [latex]a(b+c)=ab+ac[/latex]. What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variable and solve the equation.Example 2.2.F
Solve for [latex]a[/latex]. [latex]4\left(2a+3\right)=28[/latex]Answer: Apply the distributive property to expand [latex]4\left(2a+3\right)[/latex] to [latex]8a+12[/latex]
[latex]\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}[/latex]
Subtract 12 from both sides to isolate the variable term.[latex]\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}[/latex]
Divide both terms by 8 to get a coefficient of 1.[latex]\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\\a\,=\,\,2\end{array}[/latex]
Answer
[latex]a=2[/latex]Example 2.2.G
Solve for [latex]t[/latex]. [latex]2\left(4t-5\right)=-3\left(2t+1\right)[/latex]Answer: Apply the distributive property to expand [latex]2\left(4t-5\right)[/latex] to [latex]8t-10[/latex] and [latex]-3\left(2t+1\right)[/latex] to[latex]-6t-3[/latex]. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
[latex]\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Add [latex]-6t[/latex] to both sides to begin combining like terms.[latex]\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}[/latex]
Add 10 to both sides of the equation to isolate t.[latex]\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}[/latex]
The last step is to divide both sides by 14 to completely isolate t.[latex]\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}\\[/latex]
Answer
[latex-display]t=\frac{1}{2}\\[/latex-display] We simplified the fraction [latex]\frac{7}{14}\\[/latex] into [latex]\frac{1}{2}\\[/latex]Example 2.2.H
Solve [latex]\frac{1}{2}x-3=2-\frac{3}{4}x[/latex] by clearing the fractions in the equation first.Answer: Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.
[latex]\begin{array}{r}\frac{1}{2}x-3=2-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\end{array}[/latex]
Use the distributive property to expand the expressions on both sides. Multiply.[latex]\begin{array}{r}4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)-4\left(-\frac{3}{4}x\right)\\\\ \frac{4}{2}x-12=8-\frac{12}{4}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}[/latex]
Add 3x to both sides to move the variable terms to only one side. Add 12 to both sides to move the variable terms to only one side.[latex]\begin{array}{r}2x-12=8-3x\, \\\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Add 12 to both sides to move the constant terms to the other side.[latex]\begin{array}{r}5x-12=8\,\,\\ \underline{\,\,\,\,\,\,+12\,+12} \\5x=20\end{array}[/latex]
Divide to isolate the variable.[latex]\begin{array}{r}\underline{5x}=\underline{5}\\ 5\,\,\,\,\,\,\,\,\,5\\ x=4\end{array}[/latex]
Answer
[latex]x=4[/latex]Example 2.2.I
Solve [latex]3y+10.5=6.5+2.5y[/latex] by clearing the decimals in the equation first.Answer: Since the smallest decimal place represented in the equation is 0.10, we want to multiply by 10 to make 1.0 and clear the decimals from the equation.
[latex]\begin{array}{r}3y+10.5=6.5+2.5y\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 10\left(3y+10.5\right)=10\left(6.5+2.5y\right)\end{array}[/latex]
Use the distributive property to expand the expressions on both sides.[latex]\begin{array}{r}10\left(3y\right)+10\left(10.5\right)=10\left(6.5\right)+10\left(2.5y\right)\end{array}\\[/latex]
Multiply.[latex]30y+105=65+25y[/latex]
Move the smaller variable term, [latex]25y[/latex], by subtracting it from both sides.[latex]\begin{array}{r}30y+105=65+25y\,\,\\ \underline{-25y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25y} \\5y+105=65\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Subtract 105 from both sides to isolate the term with the variable.[latex]\begin{array}{r}5y+105=65\,\,\,\\ \underline{\,\,\,\,\,\,-105\,-105} \\5y=-40\end{array}[/latex]
Divide both sides by 5 to isolate the y.[latex]\begin{array}{l}\underline{5y}=\underline{-40}\\ 5\,\,\,\,\,\,\,\,\,\,\,\,\,5\\ \,\,\,x=-8\end{array}[/latex]
Answer
[latex]x=-8[/latex]Solving Multi-Step Equations
1. (Optional) Multiply to clear any fractions or decimals. 2. Simplify each side by clearing parentheses and combining like terms. 3. Add or subtract to isolate the variable term—you may have to move a term with the variable. 4. Multiply or divide to isolate the variable. 5. Check the solution.2.2.4 Classify Solutions to Linear Equations
There are three cases that can come up as we are solving linear equations. We have already seen one, where an equation has one solution. Sometimes we come across equations that don't have any solutions, and even some that have an infinite number of solutions. The case where an equation has no solution is illustrated in the next examples.2.2.5 Equations with no solutions
Example 2.2.J
Solve for x. [latex]12+2x–8=7x+5–5x[/latex]Answer: Combine like terms on both sides of the equation.
[latex] \displaystyle \begin{array}{l}12+2x-8=7x+5-5x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\end{array}[/latex]
Isolate the x term by subtracting 2x from both sides.[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,2x+4=2x+5\\\,\,\,\,\,\,\,\,\underline{-2x\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4= \,5\end{array}[/latex]
This false statement implies there are no solutions to this equation. Sometimes, we say the solution does not exist, or DNE for short.Think About It
Try solving these equations. How many steps do you need to take before you can tell whether the equation has no solution or one solution? a) Solve [latex]8y=3(y+4)+y[/latex] Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]Answer:
Solve [latex]8y=3(y+4)+y[/latex]
First, distribute the 3 into the parentheses on the right-hand side.[latex]8y=3(y+4)+y=8y=3y+12+y[/latex]
Next, begin combining like terms.[latex]8y=3y+12+y = 8y=4y+12[/latex]
Now move the variable terms to one side. Moving the [latex]4y[/latex] will help avoid a negative sign.[latex]\begin{array}{l}\,\,\,\,8y=4y+12\\\underline{-4y\,\,-4y}\\\,\,\,\,4y=12\end{array}[/latex]
Now, divide each side by [latex]4y[/latex].[latex]\begin{array}{c}\frac{4y}{4}=\frac{12}{4}\\y=3\end{array}[/latex]
Because we were able to isolate y on one side and a number on the other side, we have one solution to this equation. b) Solve [latex]2\left(3x-5\right)-4x=2x+7[/latex] Use the textbox below to record how many steps you think it will take before you can tell whether there is no solution or one solution. [practice-area rows="1"][/practice-area]Answer: Solve [latex]2\left(3x-5\right)-4x=2x+7[/latex]. First, distribute the 2 into the parentheses on the left-hand side.
[latex]\begin{array}{r}2\left(3x-5\right)-4x=2x+7\\6x-10-4x=2x+7\end{array}[/latex]
Now begin simplifying. You can combine the x terms on the left-hand side.[latex]\begin{array}{r}6x-10-4x=2x+7\\2x-10=2x+7\end{array}[/latex]
Now, take a moment to ponder this equation. It says that [latex]2x-10[/latex] is equal to [latex]2x+7[/latex]. Can some number times two minus 10 be equal to that same number times two plus seven? Let's pretend [latex]x=3[/latex]. Is it true that [latex]2\left(3\right)-10=-4[/latex] is equal to [latex]2\left(3\right)+7=13[/latex]. NO! We don't even really need to continue solving the equation, but we can just to be thorough. Add [latex]10[/latex] to both sides.[latex]\begin{array}{r}2x-10=2x+7\,\,\\\,\,\underline{+10\,\,\,\,\,\,\,\,\,\,\,+10}\\2x=2x+17\end{array}[/latex]
Now move [latex]2x[/latex] from the right hand side to combine like terms.[latex]\begin{array}{l}\,\,\,\,\,2x=2x+17\\\,\,\underline{-2x\,\,-2x}\\\,\,\,\,\,\,\,0=17\end{array}[/latex]
We know that [latex]0\text{ and }17[/latex] are not equal, so there is no number that x could be to make this equation true. This false statement implies there are no solutions to this equation, or DNE (does not exist) for short.Algebraic Equations with an Infinite Number of Solutions
You have seen that if an equation has no solution, you end up with a false statement instead of a value for x. It is possible to have an equation where any value for x will provide a solution to the equation. In the example below, notice how combining the terms [latex]5x[/latex] and [latex]-4x[/latex] on the left leaves us with an equation with exactly the same terms on both sides of the equal sign.Example 2.2.K
Solve for x. [latex]5x+3–4x=3+x[/latex]Answer: Combine like terms on both sides of the equation.
[latex] \displaystyle \begin{array}{r}5x+3-4x=3+x\\x+3=3+x\end{array}[/latex]
Isolate the x term by subtracting x from both sides.[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,x+3=3+x\\\,\,\,\,\,\,\,\,\underline{\,-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,=\,\,3\end{array}[/latex]
This true statement implies there are an infinite number of solutions to this equation, or we can also write the solution as "All Real Numbers"Example 2.2.L
Solve for x. [latex]3\left(2x-5\right)=6x-15[/latex]Answer: Distribute the 3 through the parentheses on the left-hand side.
[latex] \begin{array}{r}3\left(2x-5\right)=6x-15\\6x-15=6x-15\end{array}[/latex]
Wait! This looks just like the previous example. You have the same expression on both sides of an equal sign. No matter what number you choose for x, you will have a true statement. We can finish the algebra:[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,6x-15=6x-15\\\,\,\,\,\,\,\,\,\underline{\,-6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6x\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-15\,\,=\,\,-15\end{array}[/latex]
This true statement implies there are an infinite number of solutions to this equation.2.2.6 Absolute value equations with no solutions
As we are solving absolute value equations it is important to be aware of special cases. An absolute value is defined as the distance from 0 on a number line, so it must be a positive number. When an absolute value expression is equal to a negative number, we say the equation has no solution, or DNE. Notice how this happens in the next two examples.Example 2.2.M
Solve for x. [latex]7+\left|2x-5\right|=4[/latex]Answer: Notice absolute value is not alone. Subtract [latex]7[/latex] from each side to isolate the absolute value.
[latex]\begin{array}{r}7+\left|2x-5\right|=4\,\,\,\,\\\underline{\,-7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-7\,}\\\left|2x-5\right|=-3\end{array}[/latex]
Result of absolute value is negative! The result of an absolute value must always be positive, so we say there is no solution to this equation, or DNE.Example 2.2.N
Solve for x. [latex]-\frac{1}{2}\left|x+3\right|=6[/latex]Answer: Notice absolute value is not alone, multiply both sides by the reciprocal of [latex]-\frac{1}{2}[/latex], which is [latex]-2[/latex].
[latex]\begin{array}{r}-\frac{1}{2}\left|x+3\right|=6\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\left(-2\right)-\frac{1}{2}\left|x+3\right|=\left(-2\right)6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left|x+3\right|=-12\,\,\,\,\,\end{array}[/latex]
Again, we have a result where an absolute value is negative! There is no solution to this equation, or DNE.Summary
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques. If your multi-step equation has an absolute value, you will need to solve two equations, sometimes isolating the absolute value expression first. We have seen that solutions to equations can fall into three categories:- One solution
- No solution, DNE (does not exist)
- Many solutions, also called infinitely many solutions or All Real Numbers
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- Screenshot: Steps With an End In Sight. Provided by: Lumen Learning License: CC BY: Attribution.
- Solving Two Step Equations (Basic). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solving an Equation that Requires Combining Like Terms. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solve an Equation with Variable on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solving an Equation with One Set of Parentheses. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solving an Equation with Parentheses on Both Sides. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solving an Equation with Fractions (Clear Fractions). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Solving an Equation with Decimals (Clear Decimals). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
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- Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- Ex 4: Solving Absolute Value Equations (Requires Isolating Abs. Value). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.