unit 2 homework 6 literal equations

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2.3: Literal equations

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• Page ID 116767

• Darlene Diaz
• Santiago Canyon College via ASCCC Open Educational Resources Initiative

A literal equation is synonymous with a formula and similar to solving general linear equations because we apply the same method. We say, methods never change, just the problems. The only difference is we have several variables in the equation and we will attempt to solve for one specific variable of the formula. For example, we may have a formula such as \(A = πr^2 +πrs\), the formula for surface area of a right circular cone, and we may be interested in solving for the variable \(s\). This means we want to isolate the variable \(s\) so the equation has s isolated on one side, and everything else on the other. This looks like \[s=\frac{A-\pi r^2}{\pi r}\nonumber\]

This second equation gives the same information as the first, meaning they are algebraically equivalent. However, the original formula gives area, while the other gives \(s\), the slant height of the cone. In this section, we discuss the process in which we start from the first equation and result in the second equation.

Example \(\PageIndex{1}\)

Let’s take a look at these two examples below, side by side. The left equation is a familiar one-step equation and the right equation is also a one-step equation, this time a literal equation (or formula).

\[\begin{array}{rcl} 3x=12 & wx=z &\text{Both have coefficients} \\ &&\text{Multiply by the reciprocal of }3\text{ and }w\text{, respectively} \\ \color{blue}{\frac{1}{3}}\color{black}{}\cdot 3x=\color{blue}{\frac{1}{3}}\color{black}{}\cdot 12 & \color{blue}{\frac{1}{w}}\color{black}{}\cdot wx=z\cdot\color{blue}{\frac{1}{w}}\color{black}{}&\text{Simplify} \\ x=4& x=\frac{z}{w}&\text{Solution}\end{array}\nonumber\]

We used the same process for solving \(3x = 12\) for \(x\) as we did for solving \(wx = z\) for \(x\). Because we are solving for \(x\), we treat all the other variables the same way we would treat numbers or coefficients. Thus, we applied the multiplication property and multiplied by the reciprocal of \(3\) and \(w\) to isolate \(x\).

Solving for a Variable with One and Two-Step Equations

Example \(\pageindex{2}\).

Solve the equation \(m + n = p\) for \(n\).

\[\begin{array}{rl} m+n=p&\text{Add the opposite of }m \\ m+n+\color{blue}{(-m)}\color{black}{}=p+\color{blue}{(-m)}\color{black}{}&\text{Simplify} \\ n=p-m&\text{Solution}\end{array}\nonumber\]

Since \(p\) and \(m\) are not like terms, they cannot be combined. Hence, \(n = p − m\).

Example \(\PageIndex{3}\)

Solve the equation \(a(x − y) = b\) for \(x\).

\[\begin{array}{rl} a(x-y)=b&\text{Distribute} \\ ax-ay=b&\text{Add the opposite of }ay \\ ax+ay+\color{blue}{(-ay)}\color{black}{}=b+\color{blue}{(-ay)}\color{black}{}&\text{Simplify} \\ ax=b-ay &\text{Isolate }x\text{ by multiplying by the reciprocal of }a \\ \color{blue}{\frac{1}{a}}\color{black}{}\cdot ax=(b-ay)\cdot\color{blue}{\frac{1}{a}}\color{black}{}&\text{Simplify} \\ x=\frac{b-ay}{a}&\text{Solution}\end{array}\nonumber\]

Equivalently, \(x\) can be written as \(\frac{b}{a} − y\) by simplifying the fraction. However, it is common practice to leave it as one fraction.

Example \(\PageIndex{4}\)

Solve the equation \(y = mx + b\) for \(m\).

\[\begin{array}{rl}y=mx+b&\text{Isolate the variable term by adding the opposite of }b \\ y+\color{blue}{(-b)}\color{black}{}=mx+b+\color{blue}{(-b)}\color{black}{}&\text{Simplify} \\ y-b=mx&\text{Isolate }m\text{ by multiplying by the reciprocal of }x \\ \color{blue}{\frac{1}{x}}\color{black}{}\cdot (y-b)=mx\cdot\color{blue}{\frac{1}{x}}\color{black}{}&\text{Simplify} \\ \frac{y-b}{x}=m&\text{Rewrite with }m\text{ on the left side} \\ m=\frac{y-b}{x}&\text{Solution}\end{array}\nonumber\]

Solving for a Variable in Multiple Steps

Example \(\pageindex{5}\).

Solve the equation \(A = πr^2 + πrs\) for \(s\). This should remind you of the equation in the beginning of the section.

\[\begin{array}{rl}A=\pi r^2+\pi rs&\text{Isolate the variable term by adding the opposite of }\pi r^2 \\ A+\color{blue}{(-\pi r^2)}\color{black}{}=\pi r^2+\pi rs+\color{blue}{(-\pi r^2)}\color{black}{}&\text{Simplify} \\ A-\pi r^2=\pi rs&\text{Isolate }s\text{ by multiplying by the reciprocal of }\pi r \\ \color{blue}{\frac{1}{\pi r}}\color{black}{}\cdot(A-\pi r^2)=\pi rs\cdot\color{blue}{\frac{1}{\pi r}}\color{black}{}&\text{Simplify} \\ \frac{A-\pi r^2}{\pi r}=s&\text{Rewrite with }s\text{ on the left side} \\ s=\frac{A-\pi r^2}{\pi r}&\text{Solution}\end{array}\nonumber\]

Solving for a Variable with Fractions

Formulas often include fractions and we can solve with the same method as used previously. First, identify the LCD, and then multiply each term by the LCD. After we clear denominators, we obtain a general equation and solve as usual.

Example \(\PageIndex{6}\)

Solve the equation \(h = \frac{2m}{n}\) for \(m\).

\[\begin{array}{rl}h=\frac{2m}{n}&\text{Multiply by the LCD=}n \\ \color{blue}{n}\color{black}{}\cdot h=\frac{2m}{n}\cdot\color{blue}{n}\color{black}{}&\text{Simplify} \\ nh=2m&\text{Multiply by the reciprocal of }2 \\ \color{blue}{\frac{1}{2}}\color{black}{}\cdot nh=2m\cdot\color{blue}{\frac{1}{2}}\color{black}{}&\text{Simplify} \\ \frac{nh}{2}=m&\text{Rewrite with }m\text{ on the left side} \\ m=\frac{nh}{2}&\text{Solution}\end{array}\nonumber\]

Example \(\PageIndex{7}\)

Solve the equation \(\frac{a}{b}+\frac{c}{b}=e\) for \(a\).

\[\begin{array}{rl}\frac{a}{b}+\frac{c}{b}=e&\text{Multiply each term by the LCD=}b \\ \color{blue}{b}\color{black}{}\cdot\frac{a}{b}+\color{blue}{b}\color{black}{}\cdot\frac{c}{b}=e\cdot\color{blue}{b}\color{black}{}&\text{Simplify} \\ a+c=eb&\text{Add the opposite of }c \\ a+c+\color{blue}{(-c)}\color{black}{}=eb+\color{blue}{(-c)}\color{black}{}&\text{Simplify} \\ a=eb-c&\text{Solution}\end{array}\nonumber\]

Example \(\PageIndex{8}\)

Solve the equation \(a=\frac{A}{2-b}\) for \(b\).

\[\begin{array}{rl}a=\frac{A}{2-b}&\text{Multiply each term by the LCD}=(2-b) \\ \color{blue}{(2-b)}\color{black}{}\cdot a=\frac{A}{2-b}\cdot\color{blue}{(2-b)}\color{black}{}&\text{Simplify} \\ a(2-b)=A&\text{Distribute} \\ 2a-2b=A&\text{Isolate the variable term by adding the opposite of }2a \\ 2a-2b+\color{blue}{(-2a)}\color{black}{}=A+\color{blue}{(-2a)}\color{black}{}&\text{Simplify} \\ -2b=A-2a&\text{Multiply by the reciprocal of }-2 \\ \color{blue}{-\frac{1}{2}}\color{black}{}\cdot -2b=(A-2a)\cdot\color{blue}{-\frac{1}{2}}\color{black}{}&\text{Simplify} \\ b=-\frac{(A-2a)}{2}&\text{Distribute the negative} \\ b=\frac{-A+2a}{2}&\text{Solution}\end{array}\nonumber\]

Note, we could also write the solution as \(b = \frac{2a − A}{2}\), where the positive term is written first in the numerator. It’s not necessary, but for aesthetic reasons, we can write \(b\) this way.

The father of algebra, Persian mathematician, Muhammad ibn Musa Khwarizmi , introduced the fundamental idea of balancing by subtracting the same term from the other side of the equation. He called this process al-jabr , which later became the world Algebra.

Literal Equations Homework

Solve each of the following equations for the indicated variable.

Exercise \(\PageIndex{1}\)

\(ab = c\) for \(b\)

Exercise \(\PageIndex{2}\)

\(\frac{f}{g}x = b\) for \(x\)

Exercise \(\PageIndex{3}\)

\(3x = \frac{a}{b}\) for \(x\)

Exercise \(\PageIndex{4}\)

\(E = mc^2\) for \(m\)

Exercise \(\PageIndex{5}\)

\(V = \frac{4}{3} πr^3\) for \(π\)

Exercise \(\PageIndex{6}\)

\(a + c = b\) for \(c\)

Exercise \(\PageIndex{7}\)

\(c = \frac{4y}{m + n}\) for \(y\)

Exercise \(\PageIndex{8}\)

\(V = \frac{πDn}{12}\) for \(D\)

Exercise \(\PageIndex{9}\)

\(P = n(p − c)\) for \(n\)

Exercise \(\PageIndex{10}\)

\(T = \frac{D − d}{L}\) for \(D\)

Exercise \(\PageIndex{11}\)

\(L = L_0(1 + at)\) for \(L_0\)

Exercise \(\PageIndex{12}\)

\(2m + p = 4m + q\) for \(m\)

Exercise \(\PageIndex{13}\)

\(\frac{k − m}{r} = q\) for \(k\)

Exercise \(\PageIndex{14}\)

\(h = vt − 16t^2\) for \(v\)

Exercise \(\PageIndex{15}\)

\(Q_1 = P(Q_2 − Q_1)\) for \(Q_2\)

Exercise \(\PageIndex{16}\)

\(R = \frac{kA(T_1 + T_2)}{d}\) for \(T_1\)

Exercise \(\PageIndex{17}\)

\(ax + b = c\) for \(a\)

Exercise \(\PageIndex{18}\)

\(lwh = V\) for \(w\)

Exercise \(\PageIndex{19}\)

\(\frac{1}{a} + b = \frac{c}{a}\) for \(a\)

Exercise \(\PageIndex{20}\)

\(at − bw = s\) for \(t\)

Exercise \(\PageIndex{21}\)

\(ax + bx = c\) for \(a\)

Exercise \(\PageIndex{22}\)

\(x + 5y = 3\) for \(y\)

Exercise \(\PageIndex{23}\)

\(3x + 2y = 7\) for \(y\)

Exercise \(\PageIndex{24}\)

\(5a − 7b = 4\) for \(b\)

Exercise \(\PageIndex{25}\)

\(4x − 5y = 8\) for \(y\)

Exercise \(\PageIndex{26}\)

\(g = \frac{h}{i}\) for \(h\)

Exercise \(\PageIndex{27}\)

\(p = \frac{3y}{q}\) for \(y\)

Exercise \(\PageIndex{28}\)

\(\frac{ym}{b} = \frac{c}{d}\) for \(y\)

Exercise \(\PageIndex{29}\)

\(DS = ds\) for \(D\)

Exercise \(\PageIndex{30}\)

\(E = \frac{mv^2}{2}\) for \(m\)

Exercise \(\PageIndex{31}\)

\(x − f = g\) for \(x\)

Exercise \(\PageIndex{32}\)

\(\frac{rs}{a − 3} = k\) for \(r\)

Exercise \(\PageIndex{33}\)

\(F = k(R − L)\) for \(k\)

Exercise \(\PageIndex{34}\)

\(S = L + 2B\) for \(L\)

Exercise \(\PageIndex{35}\)

\(I = \frac{E_a − E_q}{R}\) for \(E_a\)

Exercise \(\PageIndex{36}\)

\(ax + b = c\) for \(x\)

Exercise \(\PageIndex{37}\)

\(q = 6(L − p)\) for \(L\)

Exercise \(\PageIndex{38}\)

\(R = aT + b\) for \(T\)

Exercise \(\PageIndex{39}\)

\(S = πrh + πr^2\) for \(h\)

Exercise \(\PageIndex{40}\)

\(L = π(r_1 + r_2) + 2d\) for \(r_1\)

Exercise \(\PageIndex{41}\)

\(P = \frac{V_1(V_2 − V_1)}{g}\) for \(V_2\)

Exercise \(\PageIndex{42}\)

\(rt = d\) for \(r\)

Exercise \(\PageIndex{43}\)

\(V = \frac{πr^2h}{3}\) for \(h\)

Exercise \(\PageIndex{44}\)

\(\frac{1}{a} + b = \frac{c}{a}\) for \(b\)

Exercise \(\PageIndex{45}\)

\(at − bw = s\) for \(w\)

Exercise \(\PageIndex{46}\)

\(x + 5y = 3\) for \(x\)

Exercise \(\PageIndex{47}\)

\(3x + 2y = 7\) for \(x\)

Exercise \(\PageIndex{48}\)

\(5a − 7b = 4\) for \(a\)

Exercise \(\PageIndex{49}\)

\(4x − 5y = 8\) for \(x\)

Exercise \(\PageIndex{50}\)

\(C = \frac{5}{9} (F − 32)\) for \(F\)

Ms. Lartz's Classroom

Oxford prep, math teacher, unit 2: solving equations and inequalities, quick links:, unit 2 isn set up, instructions for set-up:, 1. sign up for deltamath.com. use the code 258021., 2. add pages 23-25 to your table of contents (see slideshow to the right)., 3. tape or glue inserts to pages 23-25., 4. skip page 26., 4. add inverse operations to your table of contents and head your page (27- scroll down to see the notes)., 5. complete the inverse operations insert (yellow) and put it on page 27., 5. complete the inequalities investigation (pink)., 6. add inequalities investigation to your table of contents and head your page (28)..

1. Inverse Operations

Delta math:,    m-linear equations: one-step equations (type 1),    m-linear equations: one-step equations (t ype 2),    m-linear equations: one-step equations (t ype 3),    m-linear equations: one-step equations (t ype 4), mathletics: ,    .

2. Inequalities Investigation

   m-inequalities: numerical inequalities (true or false).

You should complete this activity.

3. Graphing and Types of Solutions

Homework keys: pages 1-2,   page 3,    a-linear inequalities: one point inequalities from number line.

4. Solving Equations and Inequalities

Homework keys: pages 4-5 ,  page 6,   pages 8-10,   page 11,    m -inequalities : linear inequalities and number line (level 1),    a-linear inequalities: linear inequalities (level 1),    a-linear inequalities: linear inequalities (level 2),    m-linear equations: linear equations w/ distribution (lev 1),    m-linear equations: linear equations w/ distribution (lev  2), remediation work:, 1. scavenger hunt- complete your work in order based on the directions once you find your answer to the equation. you must show all your work, check your answer and graph your answer to receive credit.,           link to stations, 2. complete numbers 1-7 on this worksheet..

5. Compound Inequalities

Homework key:.

Describe your image

6. Literal Equations

Remediation: ,    a-literal equations: single step literal equations (level 1),    a -literal  equations: single step literal equations (level 2),    a -literal  equations: multi-step literal equations (level 2).

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

+ Unit 1: Fundamentals of Algebra

+ Unit 2: Solving Equations and Inequalities

+ Unit 3: Linear Graphs and Inequalities

+ Unit 4: Systems of Equations and Inequalities

+ Unit 5: Descriptive Statistics

+ Unit 6: Relations and Functions

+ Unit 7: Sequences and Exponential Functions

+ Unit 8: Quadratic Expressions and Equations

+ Unit 9: Quadratic Functions

+ Unit 10: Geometry

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Friday: Math Workshops

Literal Equations

Solving literal equations.

Literal equations , simply put, are equations containing two or more variables. Your goal is to solve for just one variable with respect to others. If you know how to solve regular equations, then I guarantee you that solving literal equations will be a breeze.

What is a Literal Equation?

A literal equation is an equation that involves more than one variable. More so, a variable or “literal” is a math symbol that represents an arbitrary value or number. The letters in the alphabet are usually used to represent variables such as [latex]a[/latex], [latex]b[/latex], [latex]c[/latex], [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex]. To solve a literal equation means to express one variable with respect to the other variables in the equation.

Key Strategy to Solve Literal Equations

The “heart” of solving a literal equation is to isolate or keep by itself a certain variable on one side of the equation (either left or right) and the rest on the opposite side.

If you know how to solve regular One-Step Equations , Two-Step Equations , and Multi-Step Equations , the process of solving literal equations is very similar.

Therefore, you should not be intimidated by literal equations because you may have already the skills to tackle them. It’s just a matter of practice and familiarization.

So, the key idea looks like this. Notice that the variable that you want to solve is isolated on one side of the equation. In this case, it is on the left side. Observe that the variable [latex]\color{red}\Large{x}[/latex] is by itself on one side of the equation while the rest are on the opposite side.

Let’s go over some examples!

An Example of One-Step Literal Equation

Example 1 : Solve for [latex]s[/latex] in the literal equation [latex]P = 4s[/latex].

Remember this formula? This is the perimeter of a square where [latex]P[/latex] stands for perimeter and [latex]s[/latex] stands for the measure of one side of a square. Thus, to get the perimeter of a square we have [latex]P = s + s + s + s = 4s[/latex].

To solve for [latex]s[/latex], we need to get rid of the coefficient [latex]4[/latex] which is multiplying [latex]s[/latex].  The inverse of multiplication is division so that’s why we should divide both sides by [latex]4[/latex]!

• We can isolate variable [latex]s[/latex] on the right side.

• Divide both sides by [latex]4[/latex].

• Simplify. Now that the variable [latex]\color{red}\Large{s}[/latex] is alone on the right side of the equation, we are done!

Examples of Two-Step Literal Equations

Example 2 : Solve for [latex]L[/latex] in the literal equation [latex]P = 2L + 2W[/latex].

The literal equation mentioned above is also the formula to get the perimeter of a rectangle, where: [latex]P[/latex] = perimeter, [latex]L[/latex] = length, and [latex]W[/latex] = width. It is possible to isolate the variable [latex]L[/latex] on the right side. However, why not flip the equation around so we can keep the variable [latex]L[/latex] alone on the left?  Well, sounds like a plan!

Don’t be intimidated by how it looks. Just focus on the things you want to do, that is to solve for [latex]L[/latex] and the rest of the steps will follow.

• We want [latex]L[/latex] solved, right?

• Flip around the equation to isolate the variable on the left side.

• Subtract both sides by [latex]2W[/latex].

• Divide both sides by [latex]2[/latex].

• Simplify, now [latex]L[/latex] stands alone – solved!

Example 3 : Solve for [latex]x[/latex] in the literal equation below.

What makes this literal equation interesting is that we are going to isolate a variable that is part of the numerator of a fraction. I’m not sure if you remember that whenever you see something like this, try to get rid of the denominator first. This makes the entire solving process a lot simpler.

Since that denominator [latex]3[/latex] is dividing the expression “[latex]x − y[/latex]”, the opposite operation that can undo it is multiplication. It makes sense to multiply both sides by [latex]3[/latex] first, then add by “[latex]y[/latex]” to keep the [latex]x[/latex] by itself. Not too bad, right?

• Okay, we want to solve for [latex]x[/latex]. Let’s isolate it on the right side.

• Start by multiplying both sides by [latex]3[/latex] which is the denominator of the fraction.

• Simplify. That’s good, the denominator is now gone.

• Add both sides by [latex]y[/latex]. That’s the only way to eliminate the [latex]− y[/latex] on the right side.

• Simplify, [latex]x[/latex] is now happily by itself. Done!

Examples of Multi-Step Literal Equations

Example 4 : Solve for [latex]C[/latex] in the literal equation below.

This is the formula used to convert the measure of temperature in the Celsius unit to the Fahrenheit scale. Notice that to find the value of [latex]F[/latex] (Fahrenheit), we need to plug in some value of [latex]C[/latex] (Celsius).

However, can we also use the given formula to find Celsius whenever the value for Fahrenheit is given?

Absolutely yes!

This is a literal equation after all so it is possible to express [latex]C[/latex] in terms of [latex]F[/latex]. That’s what we’re going to do now…

• All eyes are on [latex]C[/latex]. The goal is to isolate it.

• We’ll get rid of [latex]32[/latex] on the right by subtracting both sides by [latex]32[/latex].

• This looks cleaner after simplification.

• Next, multiply both sides by [latex]5[/latex] to cancel out the denominator [latex]5[/latex] under [latex]9C[/latex].

• We’re getting there! I suggest not to distribute the [latex]5[/latex] into [latex](F − 32)[/latex].

• One more step, divide both sides by [latex]9[/latex] to finally isolate the variable on the right.

• That’s it! We have solved for [latex]C[/latex].

Example 5 : Solve for [latex]h[/latex] in the literal equation [latex]3h + g = 5h − hg[/latex].

This is really interesting! Some of you might think that it is impossible to isolate the variable [latex]h[/latex] since it is found pretty much in three places: one [latex]h[/latex] on the left and two on the right.  Well, don’t give up yet! Let me show you a little “secret”.

Use the factoring method to pick that variable [latex]h[/latex]   out of the group. But before you could factor [latex]h[/latex] out, make sure that you move all the [latex]h[/latex]’s on one side of the equation.

Since we have two terms of [latex]h[/latex]’s on the right side, we might as well move the term [latex]3h[/latex] on the left to the other side.

• We want [latex]h[/latex] isolated, right?

• Keep all our [latex]h[/latex] terms on the right side. We can do that by subtracting both sides by [latex]3h[/latex].

• After simplification, it’s wonderful to see all our [latex]h[/latex] terms just on the right side.

• It’s obvious that the step should involve factoring [latex]h[/latex] out.

Wow, this is great! Just a single [latex]h[/latex] on the right side.

• To isolate [latex]h[/latex] by itself implies that we have to get rid of the expression [latex](2−g)[/latex].

Divide both sides by [latex](2 − g)[/latex].

• Do some cancellations on the right side.

• That is it! We have solved for [latex]h[/latex].

Example 6 : Solve for [latex]x[/latex] in the literal equation below.

The most straightforward way of solving this literal equation is to perform cross multiplication . In doing so, the denominators on both sides of the equation should disappear.

From that point, we can apply the same strategy from Example 5 to solve for [latex]x[/latex] which involves gathering all [latex]x[/latex] terms on one side of the equation and then hopefully factoring the [latex]x[/latex] out.

• In this equation, we have two [latex]x[/latex]’s on both sides of the equation. More importantly, they are located in the numerator position.

• We want the denominators gone so without any hesitation we should apply the cross multiplication technique. Then simply apply the distributive property on both sides of the equation.

• At this point, we decide where to keep or gather all our [latex]x[/latex]’s. For this example, let’s keep them on the left side.

Start by getting rid of the [latex]-5x[/latex] on the right by adding [latex]5x[/latex] on both sides.

• This is how it looks after simplification. The next step is to deal with the [latex]3xy[/latex] on the right side. We want to move it to the left as well.

• Subtract both sides by [latex]3xy[/latex]. That should keep all our [latex]x[/latex]’s on the left.

• Don’t forget to write [latex]0[/latex] on the right side!

• Now, the [latex]−6[/latex] on the left must be moved to the right side. We can do this by adding [latex]6[/latex] to both sides.

• This is getting nicer! We have all our [latex]x[/latex] terms on the left. It appears that we can factor the [latex]x[/latex] out.

• Yep! We just did!

• Finally, to solve [latex]x[/latex], we should divide both sides by the expression [latex](8 − 3y)[/latex]. Perform some cancellations.

• And we’re done!

Take a Quiz !

Literal Equations Quiz

You may also be interested in these related math lessons or tutorials:

Literal Equations Practice Problems with Answers

Solving One-Step Equations

Solving Two-Step Equations

Solving Multi-Step Equations

RWM102: Algebra

Unit 2: linear equations.

We use equations everyday without realizing it. Examples include calculating the unit price to compare the price of brands in the grocery store, converting inches into feet (or centimeters into meters), estimating how much time it will take to drive to a destination at a certain speed.

In this unit, we explore formal procedures for solving equations. After reviewing basic math rules, we apply the skills we learned in Unit 1 to simplify the sides of an equation before attempting to solve it and work with equations that contain more than one variable. Because variables represent numbers, we use the same rules to find the specific variables we are looking for.

Completing this unit should take you approximately 5 hours.

Upon successful completion of this unit, you will be able to:

• determine whether a given real number is a solution of an equation;
• simplify equations using addition and multiplication properties;
• find the solution of a given linear equation with one variable;
• determine the number of solutions of a given linear equation in one variable;
• solve a literal equation for the given variable; and
• rearrange formulas to isolate a quantity of interest.

2.1: Definition of an Equation and a Solution of an Equation

We define an equation as a statement that contains a variable, which may or may not be true, depending on the value of the variable. Solving an equation means finding the possible values of the variable that make the equation true.

Read the "Define Linear Equations in One Variable" and "Solutions to Linear Equations in One Variable" sections. Then, complete exercises 1 to 5 and check your answers.

2.2: Addition/Subtraction Property of Equations

When solving algebraic equations, we need to be aware of the properties of the types of mathematical operations we are doing. The first property we explore is the addition and subtraction property of equations.

Read up to the "Solve Equations that Require Simplification" section. Pay attention to the "Solve Equations Using the Subtraction and Addition Properties of Equality" section, which gives a good example of how the two sides of an equation must be equal. After you read, complete examples 2.2 through 2.5 and check your answers.

2.3: Multiplication/Division Property of Equations

Much like in the previous section we must use the properties of multiplication and division when solving algebraic expressions involving these types of calculations.

Read up to the "Sole Equations that Require Simplification" section. Complete examples 2.13 to 2.17.

2.4: Equations of the Form x + a = b and x − a = b

Algebraic equations can be categorized based on the form and types of operations in the equation. In the next few sections, we will explore different forms of equations.

The first form is the simplest: x + a = b or x − a = b . An example of this type of equation is: 5 + x = 8.

Watch this video for examples of these types of equations.

After you watch, complete this assessment to test yourself.

2.5: Equations of the Form ax = b and x / a = b

The next general form of equations involve multiplying or dividing the variable by a coefficient. These equations are of the form ax = b or x / a = b . An example of this type of equation is: x /2 = 6.

After you watch, complete this assessment and check your answers.

2.6: Equations of the Form ax + b = c

Often types of mathematical operations are combined in an equation. For example, multiplication can be combined with addition in an equation. An example of this type of equation is: 2 x + 1 = 11. This requires a two-step process for solving the equation.

Watch this video for examples of how to solve these types of problems in a two-step process.

2.7: Equations of the Form ax + b = cx + d

This section involves solving more complicated equations where the variable appears on both sides. We can use what we learned about combining like terms to make solving these types of equations possible.

Watch these videos to see examples of how we use like terms to solve these types of equations.

2.8: Equations with Parentheses

The last general type of linear equation we can solve are those involving parentheses. For example, we can have an equation 2(4 + x ) = 12. We need to use order of operations and the distributive property to solve this type of equation.

Watch these videos to see examples of how this type of equation can be solved.

2.9: Solving Literal Equation for One of the Variables

We can use the methods we learned in the previous sections to solve literal equations, or formulas which often have more than one variable. When a literal equation has more than one variable, we can solve for the variable of interest with respect to the other variable.

For example, consider the equation 2 a + b = 10. Here, there are two variables, a and b . If we want to solve for b , we can do so with respect to a . We can subtract 2 a from both sides to obtain: 10 − 2 a = b .

Read the section on linear literal equations. Be sure to go through the examples in detail. After read, complete the exercises for literal equations and check your answers.

We can apply these concepts to known formulas, such as formulas for area of a shape or rates.

Watch these videos for real examples of using formulas. In the first video, the formula for perimeter of a rectangle is solved for the width. In the second, a formula is used to convert between Fahrenheit and Celsius.

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Unit 2 Equations Inequalities Homework 6

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