ORCCA Solving Mixed Equations

Section 13.5 Solving Mixed Equations

Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-solve-mixed-equations" missing or not unique]
[cross-reference to target(s) "ccog-check-solutions-to-equations" missing or not unique]

In this section, we will learn to differentiate between different types of equations and recall the various methods used to solve them. Real life doesn’t come with instructions, so it is important to develop the skills in this section. One day, you might be faced with a geometry problem in your home or a rational equation in a lab, and it will be your challenge to solve the equation with some strategy.

We have solved a variety of equations throughout this book, and we covered some general equation-solving strategies in Solving Equations in General. Since then, we have covered even more topics, and it seems time to refresh ourselves on everything that we have done so far. Here is a reference guide to all of the sections that cover solving equations. We hope that this section will help pinpoint those that you need help with.

Section 2.1: Solving linear equations.
Section 2.5: Solving linear equations with more than one variable.
Sections 4.2, Section 4.3: Algebraically solving systems of linear equations.
Section 6.4: Solving equations with roots in them.
Sections 7.1, Section 7.2, Section 10.7, Section 3: Solving quadratic equations.
Section 12.5: Solving rational equations.

Figure 13.5.1. Alternative Video Lesson

Subsection 13.5.1 Types of Equations

The point of this section isn’t to compartmentalize your knowledge to help learn small pieces, it’s to put all the pieces that we’ve learned previously together and how to differentiate those pieces from one another. To do that, we need to recall the different types of equations that we have had before.

Linear Equation: This is a type of equation where the variable that we are solving for only appears with addition, subtraction, multiplication and division by constant numbers. Examples are \(7(x-2)=\frac{3}{5}x+1\) and \(rt=4t+3r-1\) where \(r\) is the variable and \(t\) is considered a constant. Use the Steps to Solve Linear Equations.
System of Linear Equations: This is a grouping of two linear equations. An example is

\begin{equation*} \left\{ \begin{alignedat}{3} y \amp {}={} \amp 3x \amp {}+{} \amp 1 \\ y \amp {}={} \amp 2x \amp {}-{} \amp 5 \end{alignedat} \right. \end{equation*}

One can use either substitution or elimination to solve these systems.
Quadratic Equation: This is a type of equation where at least one side of the equation is a quadratic function, and the other side is either constant, linear, or quadratic with a different leading coefficient. Examples are \(3x^2+2x-4=0\) and \(6(y-2)^2-1=7\text{.}\) There are several methods to solve quadratic equations including using the square root method, the quadratic formula, factoring, and completing the square.
Radical Equation: This is a type of equation where the variable is inside a root of some kind. We usually solve radical equations by isolating the radical and raising both sides to a power to cancel the radical.
Rational Equation: This is a type of equation where both sides of the equation are rational functions, although it’s possible one side is a very simple rational function like a constant function. Solving these equations involves clearing the denominators and solving the equation that remains.
Absolute Value Equation: This is a type of equation where the variable is inside absolute value bars. Solving these equations involves using a rule to convert an equation from absolute value into two separate equations without absolute values.

For all of these equation types, in this section we only concern ourselves with equations in one variable, i.e. the solution will be a number or expression rather than points. For example, the equation \(3=2x+5\) has a single solution, \(-1\text{,}\) whereas the equation \(y=2x+5\) has infinitely many solutions, all points, that make up the line with slope \(2\) and vertical intercept \((0,5)\text{.}\) The only exceptions that we will be covering are systems of linear equations, which have a point or points as solutions.

Example 13.5.2.

Identify the type of equation as linear, a system of linear equations, quadratic, radical, rational, absolute value, or something else.

\(\displaystyle 3-\sqrt{2x-3}=x\)
\(\displaystyle 2x^2+3x=7\)
\(\displaystyle 7-2(3x-5)=x+\sqrt{2}\)
\(\displaystyle \frac{1}{x-2}+\frac{x}{x^2-4}=\frac{3}{x+2}\)
\(\displaystyle \abs{5x-9}+2=7\)
\(\displaystyle (4x-1)^2+9=16\)
\(\displaystyle \sqrt[3]{6x-5}=2\)
\(\displaystyle 6x^2-7x=20\)
\(\displaystyle \left\{ \begin{alignedat}{4} 4x\amp+{}\amp 2y \amp={}\amp 8 \\ 3x\amp-{}\amp y \amp={}\amp 11 \\ \end{alignedat} \right.\)
\(\displaystyle 3^x+2^x=1\)

Explanation.

The equation \(3-\sqrt{2x-3}=x\) is a radical equation since the variable appears inside the radical.
The equation \(2x^2+3x=7\) is a quadratic equation since the variable is being squared (but doesn’t have any higher power).
The equation \(7-2(3x-5)=x+\sqrt{2}\) is a linear equation since the variable is only to the first power. The square root in the equation is only on the number \(2\) and not \(x\text{,}\) so it doesn’t make it a radical equation.
The equation \(\frac{1}{x-2}+\frac{x}{x^2-4}=\frac{3}{x+2}\) is a rational equation since the variable is present in a denominator.
The equation \(\abs{5x-9}+2=7\) is an absolute value equation since the variable is inside an absolute value.
The equation \((4x-1)^2+9=16\) is a quadratic equation since if we were to distribute everything out, we would have a term with \(x^2\text{.}\)
The equation \(\sqrt[3]{6x-5}=2\) is a radical equation since the variable is inside the radical.
The equation \(6x^2-7x=20\) is a quadratic equation since there is a degree-two term.
This is a system of linear equations.
The equation \(3^x+2^x=1\) is an equation type that we have not covered and is not listed above.

Subsection 13.5.2 Solving Mixed Equations

After you have identified which type of equation confronts you, the next step is to consider the methods for solving that type of equation.

Example 13.5.3.

Solve the equations using appropriate techniques.

\(\displaystyle 3-\sqrt{2x-3}=x\)
\(\displaystyle 2x^2+3x=7\)
\(\displaystyle 7-2(3x-5)=x+\sqrt{2}\)
\(\displaystyle \frac{1}{x-2}+\frac{x}{x^2-4}=\frac{3}{x+2}\)
\(\displaystyle \abs{5x-9}+2=7\)
\(\displaystyle (4x-1)^2+9=16\)
\(\displaystyle \sqrt[3]{6x-5}=2\)
\(\displaystyle 6x^2-7x=20\)
\(\displaystyle \left\{ \begin{alignedat}{4} 4x\amp+{}\amp 2y \amp={}\amp 8 \\ 3x\amp-{}\amp y \amp={}\amp 11 \\ \end{alignedat} \right.\)
\(2x^2-12x=7\) (using completing the square)

Explanation.

Since the equation \(3-\sqrt{2x-3}=x\) is a radical equation, we can isolate the radical and then square both sides to cancel the square root. After that, we will solve whatever remains.

\begin{align*} 3-\sqrt{2x-3}\amp=x\\ -\sqrt{2x-3}\amp=x-3\\ \sqrt{2x-3}\amp=-x+3\\ \left(\sqrt{2x-3}\right)^{\highlight{2}}\amp=(-x+3)^{\highlight{2}}\\ 2x-3\amp=x^2-6x+9\\ 0\amp=x^2-8x+12\\ \end{align*}
We now have a quadratic equation. We will solve by factoring.
\begin{align*} 0\amp=(x-2)(x-6) \end{align*}

\begin{align*} x-2\amp=0\amp\amp\text{or}\amp x-6=0\\ x\amp=2\amp\amp\text{or}\amp x=6 \end{align*}

Every potential solution to a radical equation should be verified to check for any “extraneous solutions”.

\begin{align*} 3-\sqrt{2(\substitute{2})-3}\amp\wonder{=}\substitute{2}\amp\amp\text{or}\amp 3-\sqrt{2(\substitute{6})-3}\amp\wonder{=}\substitute{6}\\ 3-\sqrt{1}\amp\wonder{=}2\amp\amp\text{or}\amp 3-\sqrt{9}\amp\wonder{=}6\\ 3-1\amp\confirm{=}2\amp\amp\text{or}\amp 3-3\amp\reject{=}6 \end{align*}

So the solution set is \(\{2\}\text{.}\)
Since the equation \(2x^2+3x=7\) is quadratic we should consider the square root method, the quadratic formula, factoring, and completing the square. In this case, we will start with the quadratic formula. First, note that we should rearrange the terms in equation into standard form.

\begin{align*} 2x^2+3x\amp=7\\ 2x^2+3x-7\amp=0 \end{align*}

Note that \(a=\substitute{2}\text{,}\) \(b=\substitute{3}\text{,}\) and \(c=\substitute{-7}\text{.}\)

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(\substitute{3})\pm\sqrt{(\substitute{3})^2-4(\substitute{2})(\substitute{-7})}}{2(\substitute{2})}\\ x\amp=\frac{-3\pm\sqrt{9+56}}{4}\\ x\amp=\frac{-3\pm\sqrt{65}}{4} \end{align*}

The solution set is \(\left\{\frac{-3+\sqrt{65}}{4},\frac{-3-\sqrt{65}}{4}\right\}\text{.}\)
Since the equation \(7-2(3x-5)=x+\sqrt{2}\) is a linear equation, we isolate the variable step-by-step.

\begin{align*} 7-2(3x-5)\amp=x+\sqrt{2}\\ 7-6x+10\amp=x+\sqrt{2}\\ 17-6x\amp=x+\sqrt{2}\\ 17\amp=7x+\sqrt{2}\\ 17-\sqrt{2}\amp=7x\\ \frac{17-\sqrt{2}}{7}\amp=x \end{align*}

The solution set is \(\left\{\frac{17-\sqrt{2}}{7}\right\}\text{.}\)
Since the equation \(\frac{1}{x-2}+\frac{x}{x^2-4}=\frac{3}{x+2}\) is a rational equation, we first need to cancel the denominators after factoring and finding the least common denominator.

\begin{align*} \frac{1}{x-2}+\frac{x}{x^2-4}\amp=\frac{3}{x+2}\\ \frac{1}{x-2}+\frac{x}{(x-2)(x+2)}\amp=\frac{3}{x+2}\\ \end{align*}
At this point, we note that the least common denominator is \((x-2)(x+2)\text{.}\) We need to multiply every term by this least common denominator.
\begin{align*} \frac{1}{x-2}\multiplyright{(x-2)(x+2)}+\frac{x}{(x-2)(x+2)}\multiplyright{(x-2)(x+2)}\amp=\frac{3}{x+2}\multiplyright{(x-2)(x+2)}\\ \frac{1}{\cancelhighlight{x-2}}\cdot(\cancelhighlight{x-2})(x+2)+\frac{x}{(\cancelhighlight{x-2})(\secondcancelhighlight{x+2})}\cdot(\cancelhighlight{x-2})(\secondcancelhighlight{x+2})\amp=\frac{3}{\secondcancelhighlight{x+2}}\cdot(x-2)(\secondcancelhighlight{x+2})\\ 1(x+2)+x\amp=3(x-2)\\ x+2+x\amp=3x-6\\ 2x+2\amp=3x-6\\ 8\amp=x \end{align*}

We always check solutions to rational equations to ensure we don’t have any “extraneous solutions”.

\begin{align*} \frac{1}{\substitute{8}-2}+\frac{\substitute{8}}{\substitute{8}^2-4}\amp\wonder{=}\frac{3}{\substitute{8}+2}\\ \frac{1}{6}+\frac{8}{60}\amp\wonder{=}\frac{3}{10}\\ \frac{10}{60}+\frac{8}{60}\amp\wonder{=}\frac{3}{10}\\ \frac{18}{60}\amp\confirm{=}\frac{3}{10} \end{align*}

So, the solution set is \(\{8\}\text{.}\)
Since the equation \(\abs{5x-9}+2=7\) is an absolute value equation, we will first isolate the absolute value and then use Equations with an Absolute Value Expression to solve the remaining equation.

\begin{align*} \abs{5x-9}+2\amp=7\\ \abs{5x-9}\amp=5 \end{align*}

\begin{align*} 5x-9\amp=5\amp\amp\text{or}\amp 5x-9=-5\\ 5x\amp=14\amp\amp\text{or}\amp 5x=4\\ x\amp=\frac{14}{5}\amp\amp\text{or}\amp x=\frac{4}{5} \end{align*}

The solution set is \(\left\{\frac{14}{5},\frac{4}{5}\right\}\text{.}\)
Since the equation \((4x-1)^2+9=16\) is a quadratic equation, we again have several options. Since the variable only appears once in this equation we will use the the square root method to solve.

\begin{align*} (4x-1)^2+9\amp=16\\ (4x-1)^2\amp=7 \end{align*}

\begin{align*} 4x-1\amp=\sqrt{7}\amp\amp\text{or}\amp 4x-1\amp=-\sqrt{7}\\ 4x\amp=1+\sqrt{7}\amp\amp\text{or}\amp 4x\amp=1-\sqrt{7}\\ x\amp=\frac{1+\sqrt{7}}{4}\amp\amp\text{or}\amp x\amp=\frac{1-\sqrt{7}}{4} \end{align*}

The solution set is \(\left\{\frac{1+\sqrt{7}}{4},\frac{1-\sqrt{7}}{4}\right\}\text{.}\)
Since the equation \(\sqrt[3]{6x-5}=2\) is a radical equation, we will isolate the radical (which is already done) and then raise both sides to the third power to cancel the cube root.

\begin{align*} \sqrt[3]{6x-5}\amp=2\\ \left(\sqrt[3]{6x-5}\right)^{\highlight{3}}\amp=2^{\highlight{3}}\\ 6x-5\amp=8\\ 6x\amp=13\\ x\amp=\frac{13}{6} \end{align*}

The solution set is \(\left\{\frac{13}{6}\right\}\text{.}\)
Since the equation \(6x^2-7x=20\) is a quadratic equation, we again have several options to consider. We will try factoring on this one after first converting it to standard form.

\begin{align*} 6x^2-7x\amp=20\\ 6x^2-7x-20\amp=0\\ \end{align*}
Here, \(ac=-120\) and two numbers that multiply to be \(-120\) but add to be \(-7\) are \(8\) and \(-15\text{.}\)
\begin{align*} 6x^2\highlight{{}+8x-15x}-20\amp=0\\ \left(6x^2\highlight{{}+8x}\right)+\left(\highlight{-15x}-20\right)\amp=0\\ 2x(3x+4)-5(3x+4)\amp=0\\ (2x-5)(3x+4)\amp=0 \end{align*}

\begin{align*} 2x-5\amp=0\amp\amp\text{or}\amp 3x+4\amp=0\\ x\amp=\frac{5}{2}\amp\amp\text{or}\amp x\amp=-\frac{4}{3} \end{align*}

The solution set is \(\left\{\frac{5}{2},-\frac{4}{3}\right\}\text{.}\)
Since

\begin{align*} \left\{ \begin{alignedat}{4} 4x\amp+{}\amp 2y \amp={}\amp 8 \\ 3x\amp-{}\amp y \amp={}\amp 11 \\ \end{alignedat} \right. \end{align*}

is a system of linear equations, we can either use substitution or elimination to solve. Here we will use elimination. To use elimination, we need to make one variable have equal but opposite sign in the two equations. We will accomplish this by multiplying the second equation by \(2\text{.}\)

\begin{align*} 3x-y \amp=11\\ \multiplyleft{2}(3x-y)\amp=\multiplyleft{2}11\\ 6x-2y \amp=22 \end{align*}

So our original system becomes:

\begin{align*} \left\{ \begin{alignedat}{4} 4x\amp+{}\amp 2y \amp={}\amp 8 \\ 6x\amp-{}\amp 2y \amp={}\amp 22 \\ \end{alignedat} \right. \end{align*}

Adding the sides of the equations, we get:

\begin{align*} 10x\amp=30\\ x\amp=3 \end{align*}

Now that we have found \(x\text{,}\) we can substitute that back into one of the equations to find \(y\text{.}\) We will substitute into the first equation.

\begin{align*} 4(\substitute{3})+2y\amp=8\\ 12+2y\amp=8\\ 2y\amp=-4\\ y\amp=-2 \end{align*}

So, the solution must be the point \((3,-2)\text{.}\)
Since the equation \(2x^2-10x=7\) is quadratic and we are instructed to solve by using completing the square, we should recall how to complete the square, after we have sufficiently simplified. Let’s start by dividing all of the terms by \(2\text{.}\)

\begin{align*} 2x^2-12x\amp=7\\ x^2-6x\amp=\frac{7}{2}\\ \end{align*}
Next, we need to add \(\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=\substitute{9}\) to both sides of the equation.
\begin{align*} x^2-6x+\substitute{9}\amp=\frac{7}{2}+\substitute{9}\\ (x-3)^2\amp=\frac{7}{2}+\frac{18}{2}\\ (x-3)^2\amp=\frac{25}{2}\\ x-3\amp=\pm\sqrt{\frac{25}{2}}\\ x-3\amp=\pm\frac{\sqrt{25}}{\sqrt{2}}\\ x-3\amp=\pm\frac{5}{\sqrt{2}}\\ x-3\amp=\pm\frac{5}{\sqrt{2}}\multiplyright{\frac{\sqrt{2}}{\sqrt{2}}}\\ x-3\amp=\pm\frac{5\sqrt{2}}{2}\\ x\amp=3\pm\frac{5\sqrt{2}}{2} \end{align*}

So, our solution set is \(\left\{3+\frac{5\sqrt{2}}{2},3-\frac{5\sqrt{2}}{2}\right\}\)

Reading Questions 13.5.3 Reading Questions

1.

What are all the types of equation that you know how to solve? (Don’t worry about which types you think you should know how to solve. Just try to list all the kinds of equations that you know you know.)

2.

There are two types of equation that have been covered in this book where it is especially important to verify solutions. What are those two types of equation?

3.

What are all the ways that you know of for solving a quadratic equation? (This book has covered five general methods, but answer with as many mehtods as you know you know).

Exercises 13.5.4 Exercises

\({x^{2}+5x-50} = 0\)

Exercise Group.

25.

Solve the equation.

\(\displaystyle{ \left\lvert 2 x - 2 \right\rvert = \left\lvert 3 x + 2\right\rvert }\)

26.

Solve the equation.

\(\displaystyle{ \left\lvert 2 x - 7 \right\rvert = \left\lvert 7 x + 8\right\rvert }\)

Exercise Group.

27.

Solve the equation.

\({x^{2}+14x} = {-40}\)

28.

Solve the equation.

\({x^{2}+6x} = {-5}\)

29.

Solve the equation.

\(\displaystyle{ {\frac{1}{r+3}+\frac{3}{r^{2}+3r}} = -\frac{1}{8} }\)

30.

\({4x^{2}}={-31x-55}\)

41.

Solve the equation.

\(\displaystyle{ {t} = {\sqrt{t+7}-1} }\)

42.

Solve the equation.

\(\displaystyle{ {t} = {\sqrt{t-5}+7} }\)

43.

Solve the equation.

\(\displaystyle{ {\frac{8}{x+3}-\frac{9}{x-4}}={\frac{9}{x^{2}-x-12}} }\)

44.

Solve the equation.

\(\displaystyle{ {\frac{2}{x-7}-\frac{3}{x-4}}={-\frac{1}{x^{2}-11x+28}} }\)

Exercise Group.

45.

Solve the equation by completing the square.

\({y^{2}-2y}={63}\)

46.

Solve the equation by completing the square.

\({y^{2}-4y}={45}\)

You have attempted of activities on this page.

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