A literal equation

A literal equation DEFAULT

Solving Literal Equations

Purplemath

Sometimes we are given a formula, such as something from geometry, and we need to solve for some variable other than the "standard" one. For instance, the formula for the perimeter P of a square with sides of length s is P = 4s. We might need to solve this equation for s because we have a lot of squares' perimeters, and we want to plug those perimeter values into one formula and have that formula (maybe in our graphing calculator) spit out the value for the length of each square's side. This process of solving a formula for a specified variable is called "solving literal equations".

One of the dictionary definitions of "literal" is "related to or being comprised of letters", and variables are sometimes referred to as literals. So "solving literal equations" seems to be another way of saying "taking an equation with lots of letters, and solving for one letter in particular."

At first glance, these exercises appear to be much worse than our usual solving exercises, but they really aren't that bad. We pretty much do what we've done all along for solving linear equations and other sorts of equation; the only substantial difference is that, due to all the variables, we won't be able to simplify our work as we go along, nor as much as we're used to at the end. Here's how solving literal equations works:

This is the formula for the area A of a rectangle with base b and height h. They're asking me to solve this formula for the base b.

If they'd asked me to solve 3 = 2b for b, I'd have divided both sides by 2 in order to isolate (that is, in order to get by itself, or solve for) the variable b. I'd end up with the variable b being equal to a fractional number.

In this case, I won't be able to get a simple numerical value for my answer, but I can proceed in the same way, using the same step for the same reason (namely, that it gets b by itself). So, following the same reasoning for solving this literal equation as I would have for the similar one-variable linear equation, I divide through by the "h":

The only difference between solving the literal equation above and solving the linear equations you first learned about is that I divided through by a variable instead of a number (and then I couldn't simplify, because the fraction was in letters rather than in numbers). Because we can't simplify as we go (nor, probably, at the end), it can be very important not to try to do too much in one's head. Write everything out completely; this will help you end up with the correct answers.


This equation is the "uniform rate" equation, "(distance) equals (rate) times (time)", that is used in "distance" word problems, and solving this for the specified variable works just like solving the previous equation.

The variable they want has a letter multiplied on it; to isolate the variable, I have to divide off that letter. So I'll solve for the specified variable r by dividing through by the t:


This is the formula for the perimeter P of a rectangle with length L and width w.

If they'd asked me to solve 3 = 2 + 2w for w, I'd have subtracted the "free" 2 over to the left-hand side, and then divided through by the 2 that's multiplied on the variable. I can follow the exact same steps for this equation:

Note: I've been leaving my answers at the point where I've successfully solved for the specified variable. But this means that the variable in question has been on the right-hand side of the equation. This isn't "wrong", but some people prefer to put the solved-for variable on the left-hand side of the equation. If you prefer this, then the above answer would have been written as:

Either format is fine, mathematically, as they both mean the exact same thing.


The variable I need to isolate is currently inside a fraction. I need to get rid of the denominator. To do this, I'll multiply through by the denominator's value of 2. On the left-hand side, I'll just do the simple multiplication. On the right-hand side, to help me keep things straight, I'll convert the 2 into its fractional form of 2/1.

2Q = c + d

2Qc = c + dc

2Qc = d


If they'd asked me to solve 5 = 3 / t for t, I'd have multiplied through by t, and then divided both sides by 5. Following the same reasoning and doing the same steps, I get:


This next exercise requires a little "trick" to solve it.

I need to get the variable a by itself. Currently, it's multiplied onto other stuff in two different terms. I can't combine those terms, because they have different variable parts. (The first term has no other variable, but the second term also has the variable c.) I want to divide off the stuff that's multiplied on the specified variable a, but I can't yet, because there's different stuff multiplied on it in the two different places. But what if I factor the a out front?

The "trick" came in the second line, where I factored the a out front on the right-hand side. By doing this, I created one (big, lumpy) multiplier on a, which I could then divide off.

This technique (factoring out to allow for dividing through) doesn't come up often, but it's just about guaranteed to come up in your homework once or twice, and almost-certainly on your next test, precisely because so many students don't see the "trick". So keep in mind: When you can't isolate the desired variable because it is a factor in two or more terms, collect those terms together on one side of the "equals" sign, factor out the desired variable, and then divide through by whatever is left after you factored.


  • Solve A = (½)ah – (½)bh for h

Since each of the two fractions on the right-hand side has the same denominator of 2, I'll start by multiplying through by 2 to clear the fractions. Then I'll work toward isolating the variable h.

This example used the same "trick" as the previous one. In the fourth line, I factored out the h. You should expect to need to know how to do this!


  • The area A of a sector (a pie-wedge-shaped section) of a circle with radius r and angle-measure S (in degrees) is given by Solve this equation for S.

This is a big, lumpy equation, but the solution method is the same as always. The variable I want has some other stuff multiplied onto it and divided into it; I'll divide and multiply through, respectively, to isolate what I need.


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What is the difference between a literal equation and a formula?

What is the difference between a literal equation and a formula?

A literal equation is an equation which consists primarily of letters. Formulas are an example of literal equations. Each variable in the equation “literally” represents an important part of the whole relationship expressed by the equation.

How are literal equations similar to and different from equations in one variable?

Answer Expert Verified A literal equation is, simply put, an equation that has a lot of letters or variables. The difference is literal equation have more variables, you solve for certain variable in terms of another variables while in one variable you get the direct number.

What is an example of a literal equation?

A literal equation is one that has several letters or variables. Examples include the area of a circle (A=πr2) and the formula for speed (v=Dt). If the unknown variable is in the denominator, we multiply both sides by the lowest common denominator (LCD) and then continue to solve.

How do you explain literal equations?

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 a, b, c, x, y, and z.

Why is it called a literal equation?

A literal equation is an equation where variables represent known values. Literal equations allow use to represent things like distance, time, interest, and slope as variables in an equation. Using variables instead of words is a real time-saver!

How do you solve multi step equations?

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.

How do you solve system of equations?

Here’s how it goes:

  1. Step 1: Solve one of the equations for one of the variables. Let’s solve the first equation for y:
  2. Step 2: Substitute that equation into the other equation, and solve for x.
  3. Step 3: Substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.

Is 9r 16 PI 5 a literal equation?

That is not a literal equation, because the pi symbol is not a variable, you can do 3.14 divided by 5.

Are literal equations the same as linear equations?

are both literal equations. The process is the same process that you use to solve linear equations; the only difference is that you will be working with a lot more letters, and you may not be able to simplify as much as you can with linear equations.

Does a literal equation have only one variable?

A literal equation has only one variable.

What is an equation called with more than one variable?

literal equation

What is an equation that has two or more variables?

A system of equations (also known as simultaneous equations) is a set of equations with multiple variables, solved when the values of all variables simultaneously satisfy all of the equations.

How do you solve problems with two variables?

Divide both sides of the equation to “solve for x.” Once you have the x term (or whichever variable you are using) on one side of the equation, divide both sides of the equation to get the variable alone. For example: 4x = 8 – 2y.

How do you solve an equation with two variables on both sides?

In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. Step 1: Multiply equation (1) by -5 and add it to equation (2) to form equation (3) with just one variable.

What are terms in an equation?

A term is a single mathematical expression. It may be a single number (positive or negative), a single variable ( a letter ), several variables multiplied but never added or subtracted. Some terms contain variables with a number in front of them. The number in front of a term is called a coefficient.

How do you find the number of terms in an equation?

Determine an expression of which you want to identify the terms. For example, use 3x^2 + 4y + 5. Find the number, variable or number multiplied by a variable before the first operator in the expression, starting from left to right, to identify the first term in the expression.

What are the types of equation?

Different Types of Equations

  • Quadratic Equation.
  • Linear Equation.
  • Radical Equation.
  • Exponential Equation.
  • Rational Equation.

What are terms examples?

A term can be a constant or a variable or both in an expression. In the expression, 3a + 8, 3a and 8 are terms. Here is another example, in which 5x and 7 are terms that form the expression 5x + 7. Math Games for Kids.

What are two terms in math?

Example: ax2 + bx + c A Term is either a single number or a variable, or numbers and variables multiplied together. An Expression is a group of terms (the terms are separated by + or − signs)

What is factor in math definition?

Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly.

What are factors of 18?

Factors of 18

  • Factors of 18: 1, 2, 3, 6, 9 and 18.
  • Negative Factors of 18: -1, -2, -3, -6, -9 and -18.
  • Prime Factors of 18: 2, 3.
  • Prime Factorization of 18: 2 × 3 × 3 = 2 × 32
  • Sum of Factors of 18: 39.

What’s a factor of 24?

What are the Factors of 24? The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 and its negative factors are -1, -2, -3, -4, -6, -8, -12, -24.

07/07/2021Manon WilcoxEducation

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11.10. Literal Equations and Formulas

A literal equation is one in which some or all of the constants are represented by letters.

The following is a literal equation:

a(x + b) = b(x + c)

A formula is a literal equation that relates two or more mathematical or physical quantities. These are the equations that describe the workings of the physical world. In Chap. 1 we substituted into formulas. Here we solve formulas or other literal equations for one of its quantities.

11.10.1. Solving Literal Equations and Formulas

When we solve a literal equation or formula, we cannot, of course, get a numerical answer, as we could with a numerical equation. Our object here is to isolate one of the letters on one side of the equal sign. We "solve for" one of the literal quantities.

Solve for x:

a(x + b) = b(x + c)

Solution: Our goal is to isolate x on one side of the equation. Removing parentheses, we obtain

ax + ab = bx + bc

Subtracting bx and then ab will place all of the x terms on one side of the equation.

ax − bx = bc − ab

Factoring to isolate x yields

x(a − b) = b(c − a)

Dividing by (a − b), where a ≠ b, gives us

Check: We substitute our answer into our original equation.

We solve literal equations by calculator in the same way we solved numerical equations. Select ...

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How to Solve LITERAL EQUATIONS - ALGEBRA 1 - Remote Learning

What is a literal equation example?

A literal equation is one that has several letters or variables. Examples include the area of a circle (A=πr2) and the formula for speed (v=Dt). If the unknown variable is in the denominator, we multiply both sides by the lowest common denominator (LCD) and then continue to solve.

What’s a literal equation?

A literal equation is an equation which consists primarily of letters. Formulas are an example of literal equations. Each variable in the equation “literally” represents an important part of the whole relationship expressed by the equation.

What is the most common type of literal equation?

Terms in this set (5)

  • Literal Equation. An equation that has more than one variable.
  • Formula. The most common type of literal equation.
  • A=bh/2. An example of a formula (this is the formula for area of a rectangle)
  • In formulas.
  • To solve for x.

Why is it called a literal equation?

Literal equation are equations (meaning they have an equals sign) that contain more than one distinct variable. For instance, x2+y2=1 is a literal equation because x and y both appear in the equation. On the other hand x2+x=1 is NOT a literal equation because x is the only variable that appears.

What is a non example of an equation?

* 5x+2y-3z is not an equation. * 144÷6x=12 is an example of an equation. * 27÷3>5 is not an equation.

What are examples and non examples?

Examples provide synonyms or other means of similarity with the unclear concept. Therefore, an example is always like or similar to whatever it is an example of. Nonexamples are, as you can tell, the opposite of examples. Where examples provide an instance of similarity, nonexamples provide an instance of contrast.

How you will know if the equation is a polynomial equation?

In particular, for an expression to be a polynomial term, it must contain no square roots of variables, no fractional or negative powers on the variables, and no variables in the denominators of any fractions.

What does non mean in math?

not necessarily

What is non-example?

nonexample(Noun) Example that is irrelevant to a rule or a definition already shown, used for a clearer explanation. Etymology: non- and example.

What are examples of non polynomials?

3×2 – 2x-2 is not a polynomial because it has a negative exponent. is not a polynomial because it has a variable under the square root. is not a polynomial because it has a variable in the denominator of a fraction.

What is the importance of giving the examples and non examples?

One way to teach concepts then is to give examples and non-examples of the words. When one gives an example and the opposite it allows the student to compare and contrast the two words. This helps to highlight the attributes of the word being taught that makes it different from other words.

Why are non examples important?

A Non-Example is simply helps define a new term in it’s entirety using the characteristics that are given and determine what the term is not. Non-examples don’t always have to be done in the form of a vocabulary definition. Another way to use them is through a concept sort or Venn diagram.

What are non examples of globalization?

Answer:

  • Social movements contrary to neoliberal policies.
  • Environmental movements.
  • Indigenous movements,
  • Unionists.
  • Anarchists.

How do you teach concepts?

Concept Teaching Instructional Strategy

  1. Select Big Idea concepts and determine the best approach:
  2. Clarify aims/establish a “hook” to draw students in.
  3. Proceed through the selected inductive or deductive approach using examples & nonexamples.
  4. Get students to demonstrate their understanding.

Is the formation of ideas or concepts?

Concept Formation is an inductive teaching strategy that helps students form a clear understanding of a concept (or idea) through studying a small set of examples of the concept. Concepts are the “furniture” of our minds. When these similarities are established in students’ minds, they form the concept.

What are the 3 ways in explaining a concept?

In contemporary philosophy, there are at least three prevailing ways to understand what a concept is: Concepts as mental representations, where concepts are entities that exist in the mind (mental objects) Concepts as abilities, where concepts are abilities peculiar to cognitive agents (mental states)

How do you start a concept essay?

How to Write an Essay Explaining a Concept

  1. First, what you need to do is to present the main idea. Make it a point that your audience or readers will be familiar with the given concept.
  2. Second, present your supporting topics.
  3. Lastly, provide a summary writing skills of your explanation.

How do you write a concept paragraph?

  1. Keep it Brief. In many instances, a concept statement of one sentence is probably a bit too short, but in most instances, a full page is likely too long.
  2. Explain Your Idea Clearly.
  3. Write to Your Audience.
  4. Spell Out the Benefits.

What is a concept paragraph?

The conceptual paragraph is defined as a group of rhetorically related concepts developing a generalization to form a coherent and complete unit of discourse and consisting of one or more traditional physical paragraphs.

How do you write a concept?

First, write a concept outline.

  1. That means you have the right idea for the insight, the right idea for the benefit, and the right ideas for the RTBs.
  2. Write them all down in outline form.
  3. Check to see if it all hangs together and creates a logical story.
  4. Have others read it over to see if they agree it’s basically right.
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Equation a literal

Solving for a Variable in a Formula – Literal Equations

What are Literal Equations?

The use of formulas is very common in science and engineering. The formulas are manipulated to have a variable initially on the RHS, becoming the subject of the formula on the LHS. I know you have too encountered numerous formulas in your journey of studying Algebra.

Most mathematical formulas are based on geometrical concepts.
For instance, you may have come across formulas such as area of a rectangle (A = l × w), area of a circle (A = πr2), distance formula (D = v × t), etc. These kinds of formulas are known as literal equations.

The word “literal” means “related to,” and variables are sometimes called literals. Therefore, we can define literal equations as equations that contain two or more variables.

How to Solve Literal Equations?

Solving a literal equation means taking an equation with lots of variables and solving one of the variables in particular. The procedures used to solve regular One-Step Equations, Two-Step Equations, and Multi-Step Equations are also applied to solve literal equations.

The objective of solving these equations is to isolate a given variable from an equation. The only difference while solving literal equations is that process involves several letters, and simplification of the equation is limited.

This article will step by step guide you in understanding how to solve literal equations so that you can solve literal equations yourself.

Let’s look at a couple of examples below.

Example 1

Given the area of a rectangle as A = w × h, we can manipulate the variables in the equation as illustrated below:

To isolate the width(w) to the equation’s left side, A = w × h. Swap the equation and divide both sides by the height (h).

(w × h)/h = A/h

w = A/h

To isolate h on the left side, also divide both sides by w.

(w × h)/w = A/w

h = A/w

Example 2

Consider the formula for the area of a circle: A = π r2.

To isolate the radius (r) on the left side of the equation, swap the equation and divide both sides by pi (π).

(π r2) = A/ π

r2 = A/ π

To remove the exponent from r, find the positive square root of both sides of the equation.

√ r2 = √ (A/ π)

r = √ (A/ π)

Example 3

Solve for x in the literal equation 3x + y = 5x – xy.

Isolate all variables having x on the right side by subtracting 3x from both sides of the equation.

3x – 3x + y = 5x – 3x – xy

y = 2x – xy

Factorize x out in the equation

y = x (2 – y)

Now divide both sides of the equation by 2 – y

y/(2 – y) = x (2 – y)/(2 – y)

y/(2 – y) = x

That is it!

Example 4

Given the literal formula: t = a + (n – 1) d, find the value of d when
t = 10, a = 2, n = 5.
Solution

First make d the subject of the formula and substitute the values.
d = (t – a)/ (n – 1)
Now, substitute the values of t, n and a.

d = (10 – 2)/ (5 – 1)
= 8/4
= 2

Example 5

Solve for R in the following literal equation S = 3R + 5RZ.

Solution

In this case, we need to isolate variable R, and yet it is multiplied onto other terms.

The first step is to factorize R out.

S = R (3 + 5Z)

Divide both sides by (3 + 5Z).

S/ (3 + 5Z) = R (3 + 5Z)/ (3 + 5Z)

S/ (3 + 5Z) = R

Example 6

Solve T in the following equation H = (1/4) KT– (1/4) RT.

Solution

Since the expression on the right has a 4, start by multiplying through by 4 to eliminate the fractions.

4H = [ (1/4) KT– (1/4) RT]4

4H = KT– RT.

Swap the equation and factorize T out.

T (K– R) = 4H

Divide both sides by (K– R)

T (K– R)/ (K– R) = 4H / (K– R)

T= 4H / (K– R)

That is it! We have solved for T.

Example 7

Solve for y in the following formula: 2y + 4x = 2.

Solution

Subtract both sides by 4x to isolate 2y.

2y + 4x – 4x = 2 – 4x

2y = 2 – 4x

Divide by 2.

2y/2= (2 – 4x)/2

y= (2 – 4x)/2

Simplify the equation;

y= 2/2 – 4x/2

y= 1 – 2x

And that is the answer.

Example 8

Given the formula p = 2(L+ b), Calculate the value of b when P and L are 36 and 10, respectively.
Solution

The first step is to make b the subject of the formula, and then we substitute the given values of P and L.
P = 2 (L + b)

Remove the parentheses applying the distributive property of multiplication.
P = 2L + 2b

Subtracting by 2L on both sides of the equation gives;
P – 2L= 2b

Now divide both sides by 2.
(P – 2L)/2 = 2b/2
b = (P – 2L)/2

If P= 36 and L= 10, substitute the values in the equation to get b.

b = (36 – 2 × 10)/2

b = (36 – 20)/2

b = 16/2
b = 8

Example 9

The perimeter of a rectangular is given by P = 2L + 2w, where p = perimeter, L = length and w = width. Make L the subject of the formula.

Solution

We have decided to keep L on the right side by subtracting both sides by 2w.

P – 2w = 2L + 2w- 2w

P – 2w = 2L

Divide both sides of the equation by 2.

(P – 2w)/ 2 = 2L/2

P/2 -w = L

Yep! We are done.

Example 10

Find for t in the following literal equation v = u + at.

Solution

Subtract u from both sides.
v – u = u – at – u
v – u = at
On dividing both sides by a we obtain;

(v – u)/a = at/a
t = (v – u)/a

How to solve literal equations with fractions?

Let’s understand this concept with the help of a few examples below:

Example 11

Make y the subject of the formula in the following literal equation x = (y + z)/ (y – z)
Solution

Multiply both sides by (y – z)
x = (y + z)/ (y – z)
x (y – z) = y + z
xy – xz = y + z
xy – y = z + zx
y (x – 1) = z (x + 1)
y = z (x + 1)/ (x – 1)

Example 12

Solve A in the literal equation below:

B/5 = (A – 32)/9

Solution
B/5 = (A – 32)/9
⇒ 9B/5 = A – 32
⇒ 9B/5 + 32 = A
⇒ A = 9B/5 + 32

Example 13

Given a literal formula A = P {1 + (r/100)} ⁿ. Find r when A = 1102.50, P = 1000 and n is given as 2.
Solution
A = P {1 + (r/100)} ⁿ

Divide both sides of the equation by P.

A/P = {1 + (r/100)} ⁿ

Calculate the nth root on both sides of the equation.

(A/P)1/n = {1 + (r/100)}

Subtract both sides by 1.
(A/P)1/n – 1 = r/100

Multiply both sides by 100 to eliminate the fraction.
100 {(A/P)1/n – 1} = r
To find the numerical value of r, substitute p the values of P, n and A in the equation.

r = 100 {(1102.50/1000)1/2 – 1}
= 100 {(110250/1000)1/2 – 1}
= 100 {(441/400)1/2 – 1}
= 100 [{(21/20)2}1/2 – 1]
= 100 {(21/20)2 x 1/2 – 1}

= 100 {21/20 – 1}
= 100 {(21 – 20)/20}
= 100 × 1/20
= 5

Example 14

Make d the subject of the formula Q = (c + d​)/2

Solution

Cross multiply the equation and eliminate the brackets:

Q= (c + d​)/2 => 2Q = c + d

To isolate d subtract both sides by c

2Q – c = c- c + d

2Q – c = d

d = 2Q – c. And we’re done!

Example 15

Solve for x in the following literal equation

(x -2)/ (3y – 5) = x/3

Solution

This kind of equation is having rational expression on both sides, therefore we perform cross multiplication;

(x -2)/ (3y – 5) = x/3 => 3(x-2) = x (3y – 5)

Apply the distributive property of multiplication to remove the brackets;

3x – 6 = 3xy – 5x

Let’s keep x’s on the left side.

Eliminate -5x on the right by adding 5x to both sides

3x + 5x – 6 = 3xy – 5x + 5x

8x -6 = 3xy

To keep all x’s on the left, subtract both sides by 3xy.

8x -3xy -6 = 3xy -3xy

8x – 3xy – 6 = 0

Now transfer the constant on the right side by adding both sides by 6.

8x – 3xy – 6 + 6= 0 + 6

8x – 3xy = 6

Factorize out x.

x (8x – 3y) = 6

Divide both sides by 8x-3y

x (8x – 3y)/ (8x – 3y) = 6/ (8x – 3y)

x=6/ (8x – 3y)

And that is the answer!

Practice Questions

  1. Make x the subject of the formula: y = 4x + 3.
  2. Make y the subject of: x = 2 – 5y
  3. Make y the subject of: w2 = x 2 + y2
  4. Solve for x in the following literal equation: 3(x + a) = k (x – 2)
  5. Make x the subject of the formula: ax + 3= bx + c
  6. Solve for s given the formula: a – xs = b – sy
  7. Make z the subject of the formula: 4y + 2 = z – 4
  8. Make m the subject of the formula: T – m = am/2b
  9. Make t the subject of the formula: r = a + bt2
  10. Make p the subject of the formula given t =wp2/32r

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KutaSoftware: Algebra 1 - Literal Equations Part 1

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 a, b, c, x, y, and z. 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 it. 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 \color{red}\Large{x} is by itself on one side of the equation while the rest are on the opposite side.

 the solution of a literal equation may contain variables not just constants.

Let’s go over some examples!

An Example of One-Step Literal Equation

Example 1: Solve for s in the literal equation P = 4s.

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

a blue square with a side measure of "s" which implies that its perimeter is four times the length of one side.

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

  • We can isolate variable s on the right side.
P is equal to the product of 4 and s. In short, we have P=4s.
the quotient of P and 4 is equal to the quotient of 4s and 4. We can write this as P/4=(4s)/4.
  • Simplify. Now the variable \color{red}\Large{s} is alone on the right side of the equation. We are done!
P divided by 4 equals s. We write this in an equation as P/4=s.

Examples of Two-Step Literal Equations

Example 2: Solve for L in the literal equation P = 2L + 2W.

Well, this is the formula to get the perimeter of a rectangle where: P = perimeter, L = length, and W = width. It is possible to isolate the variable L on the right side, however, why not flip the equation around so we can keep the variable L alone on the left?  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 L and the rest of the steps will follow.

P is equal to the sum of the products of 2 and L, and 2 and W. It can be written as P=2L+2W.
  • Flip around the equation to isolate the variable on the left side.
2L+2W=P
  • Subtract both sides by 2W.
2L+2W-2W=P-2W
2L=P-2W
(2L)/2=(P-2W)/2
  • Simplify, now L stands alone – solved!
L=(P-2W)/2

Example 3: Solve for x in the literal equation below.

m is equal to the ratio of the difference of x and y, and 3. we can write this as m=(x-y)/3.

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

  • Okay, we want to solve for x. Let’s isolate it on the right side.
m=(x-y)/3
  • Start by multiplying both sides by 3 which is the denominator of the fraction.
multiply 3 on both sides of the equation m=(x-y)/3.
  • Simplify. That’s good, the denominator is now gone.
3m=x-y
  • Add both sides by y. That’s the only way to eliminate that − y on the right side.
3m+y=x-y+y
  • Simplify, x is now happily by itself. Done!
3m+y=x

Examples of Multi-Step Literal Equations

Example 4: Solve for C in the literal equation below.

this is the formula to convert Celsius to Fahrenheit
thermometer

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 F (Fahrenheit), we need to plug in some value of C (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 C in terms of F. That’s what we’re going to do now…

  • All eyes on C. The goal is to isolate it.
F=(9C)/5+32
  • We’ll get rid of that 32 on the right by subtracting both sides by 32.
F-32=+32-32
  • This looks cleaner after simplification.
F-32=(9C)/5
  • Next, multiply both sides by 5 to cancel out that denominator 5 under 9C.
multiply 5 to both sides of the equation F-32=(9C)/5.
  • We’re getting there! I suggest not to distribute that 5 into (F − 32).
5(F-32)=9C
  • One more step, divide both sides by 9 to finally isolate the variable on the right.
divide both sides by 9 to get the formula to determine Celsius value when a Fahrenheit temperature is given
  • That’s it! We have solved for C.
C=(5/9)✖️(F-32)

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

This is really interesting! Some of you might think that it is impossible to isolate the variable h since it is found pretty much in three places – one h 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 out of the group. But before you could factor h out, make sure that you move all the h’s on one side of the equation.

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

  • We want h isolated, right?
3h+g=5h-hg
  • Keep all our h terms on the right side. We can do that by subtracting both sides by 3h.
3h-3h+g=5h-3h-hg
  • After simplification, it’s wonderful to see all our h terms just on the right side.
g=2h-hg
  • It’s obvious that the step should involve factoring h out.
factor out h on the right side of the equation to get g=h(2-g).

Wow, this is great! Just a single h on the right side.

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

Divide both sides by (2 − g).

divide the entire literal equation by 2-g to isolate the variable h. g/(2-g)=/(2-g).
  • Do some cancellation on the right side.
cancel out common terms which give us g/(2-g)=h(2-g)/(2-g).
  • That is it! We have solved for h.
the final solution of the given literal equation is h=g/(2-g).

Example 6: Solve for x in the literal equation below.

(x-2)/(3y-5)=x/3

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 x which involves gathering all x terms on one side of the equation and then hopefully factor the x out.

  • In this equation, we have two x’s on both sides of the equation. More importantly, they are located in the numerator position.
(x-2)/(3y-5)=x/3
  • We want the denominators gone so without any hesitation we should apply the cross multiplication technique.
3(x-2)=x(3y-5)
  • Then simply apply distributive property on both sides of the equation. If you forgot how to do it, check it out in this lesson.
3x-6=3xy-5x
  • At this point, we decide where to keep or gather all our x’s. For this example, let’s keep them on the left side.

Start by getting rid of the -5x on the right by adding both sides by 5x.

3x+5x-6=3xy-5x+5x
  • This is how it looks after simplification. Next is to deal with that 3xy on the right side. Move that to the left as well.
8x-6=3xy
  • Subtract both sides by 3xy. That should keep all our x’s on the left.
8x-3xy-6=3xy-3xy
  • Don’t forget writing that 0 on the right side!
8x-3xy-6=0
  • Now that −6 on the left must be moved to the right side by adding both sides by 6.
8x-3xy-6+6=0+6
  • This is getting nicer! We have all our x terms on the left. It appears that we can factor the x out.
8x-3xy=6
x(8-3y)=6
  • Finally, to solve x, we should divide both sides by the expression (8 − 3y). Perform some cancellation.
[x(8-3y)]/(8-3y)=6/(8-3y)
x=6/(8-3y)

Literal Equations Worksheets

Download Version 1

Download Version 2


You might be interested in:

Solving One-Step Equations

Solving Two-Step Equations

Solving Multi-Step Equations

Sours: https://www.chilimath.com/lessons/intermediate-algebra/literal-equations/

Now discussing:

Solving Literal Equations - Tutorial

How to Solve Literal Equations? Examples with Detailed Solutions

Example 1

Solve the literal equation
P = 2L + 2W
for W.

Solution to Example 1

  • Given
    P = 2L + 2W
  • we first isolate the term containing W : add -2L to both sides of the equation
    P - 2L = 2L + 2W - 2L
  • Simplify to obtain
    P - 2L = 2W
  • Divide both sides by 2 to obtain W.
    W = (P- 2L) / 2

Example 2

Solve the literal equation
H = √ ( x 2 + y 2)

for y, where H, x and y are a positive real numbers and H is greater than x and greater than y.

Solution to Example 2

  • Given
    H = √ ( x 2 + y 2)
  • Square both sides
    H 2 = x 2 + y 2
  • Add - x 2 to both sides and simplify
    H 2 - x 2 = x 2 + y 2 - x 2
    H 2 - x 2 = y 2
  • Solve for y taking the square root
    y = ~+mn~ √ (H 2 - x 2)
  • Since y is a positive real number, then y is given by
    y = + √ (H 2 - x 2)

Example 3

Express F in terms of C in the literal equation
C = (5 / 9)(F - 32)
.

Solution to Example 3


    C = (5 / 9)(F - 32)
  • Multiply both sides of the formula by 9 / 5
    (9 / 5) C = (9 / 5)(5 / 9)(F - 32)
  • and simplify
    (9 / 5) C = (F - 32)
  • Add 32 to both sides of the formula.
    (9 / 5) C + 32 = F
  • The formula F = (9 / 5) C + 32 expresses F in terms of C.

Example 4

Express y in terms of x in the literal equation
a x + b y = c , with b not equal to zero.
.

Solution to Example 4


    a x + b y = c
  • Add - a x to both sides of the equation
    a x + b y - a x = c - a x
    b y = - a x + c
  • Divide both sides by b.
    y = - (a / b) x + c / b

Exercises

Solve each of the literal equations below for the indicated variable.(see answers below).
  1. A = W L , for L.

  2. y = m x + b , for x.

  3. A = (1 / 2)(B + a) , for a.

  4. S = 2 π r h , for r.

  5. F = (9 / 5)C + 32 , for C.

  6. 1 / x = 1 / y + 1 / z , for y.

Answers to Above Exercises:


  1. L = A / W

  2. x = (y - b) / m , for m not equal to zero.

  3. a = 2 A - B

  4. r = S / (2 π h)

  5. C = (5 / 9)(F - 32)

  6. y = (x z) / (z - x) , for z not equal to x.

More References and links

Solve Equations, Systems of Equations and Inequalities.
Sours: https://www.analyzemath.com/Equations/literal_equations.html


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