Learning Objectives

By the end of this section, you will be able to:

- Solve a formula for a specific variable
- Use formulas to solve geometry applications

Before you get started, take this readiness quiz.

- Evaluate (2(x+3)) when (x=5).

If you missed this problem, review [link]. - The length of a rectangle is three less than the width. Let
*w*represent the width. Write an expression for the length of the rectangle.

If you missed this problem, review [link]. - Evaluate (frac{1}{2}bh) when (b=14) and (h=9).

If you missed this problem, review [link].

#### Solve a Formula for a Specific Variable

We have all probably worked with some geometric formulas in our study of mathematics. Formulas are used in so many fields, it is important to recognize formulas and be able to manipulate them easily.

It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. We isolate that variable on one side of the equals sign with a coefficient of one and all other variables and constants are on the other side of the equal sign.

Geometric formulas often need to be solved for another variable, too. The formula (V=frac{1}{3}πr^2h) is used to find the **volume** of a right circular cone when given the radius of the base and height. In the next example, we will solve this formula for the height.

EXAMPLE (PageIndex{1})

Solve the formula (V=frac{1}{3}πr^2h) for *h*.

**Answer**Write the formula. Remove the fraction on the right. Simplify. Divide both sides by (πr^2). We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula (h=frac{3V}{πr^2}).

EXAMPLE (PageIndex{2}):

Use the formula (A=frac{1}{2}bh) to solve for *b*.

**Answer**(b=frac{2A}{h})

EXAMPLE (PageIndex{3}):

Use the formula (A=frac{1}{2}bh) to solve for *h*.

**Answer**(h=frac{2A}{b})

In the sciences, we often need to change **temperature** from Fahrenheit to Celsius or vice versa. If you travel in a foreign country, you may want to change the Celsius temperature to the more familiar Fahrenheit temperature.

EXAMPLE (PageIndex{4})

Solve the formula (C=frac{5}{9}(F−32)) for *F*.

**Answer**Write the formula. Remove the fraction on the right. Simplify. Add 32 to both sides. We can now use the formula (F=frac{9}{5}C+32) to find the Fahrenheit temperature when we know the Celsius temperature.

EXAMPLE (PageIndex{5}):

Solve the formula (F=frac{9}{5}C+32) for *C*.

**Answer**(C=frac{5}{9}(F−32))

EXAMPLE (PageIndex{6}):

Solve the formula (A=frac{1}{2}h(b+B)) for *b*.

**Answer**(b=frac{2A−Bh}{h})

The next example uses the formula for the **surface area** of a right cylinder.

EXAMPLE (PageIndex{8}):

Solve the formula (A=P+Prt) for (t).

**Answer**(t=frac{A−P}{Pr})

EXAMPLE (PageIndex{9}):

Solve the formula (A=P+Prt) for (r).

**Answer**(r=frac{A−P}{Pt})

Sometimes we might be given an equation that is solved for (y) and need to solve it for (x), or vice versa. In the following example, we’re given an equation with both (x) and (y) on the same side and we’ll solve it for (y).

Solve the formula (4x+7y=9) for (y).

**Answer**(y=frac{9−4x}{7})

EXAMPLE (PageIndex{12})

Solve the formula (5x+8y=1) for (y).

**Answer**(y=frac{1−5x}{8})

#### Use Formulas to Solve Geometry Applications

In this objective we will use some common geometry formulas. We will adapt our problem solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

SOLVE GEOMETRY APPLICATIONS.

**Read**the problem and make sure all the words and ideas are understood.**Identify**what you are looking for.**Name**what we are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.**Translate**into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.**Solve**the equation using good algebra techniques.**Check**the answer in the problem and make sure it makes sense.**Answer**the question with a complete sentence.

When we solve geometry applications, we often have to use some of the properties of the figures. We will review those properties as needed.

The next example involves the area of a triangle. The** area** of a triangle is one-half the base times the height. We can write this as (A=frac{1}{2}bh), where (b) = length of the base and (h) = height.

EXAMPLE (PageIndex{13})

The area of a triangular painting is (126) square inches. The base is (18) inches. What is the height?

**Answer****Step 1. Read**the problem.**Step 2. Identify**what you are looking for.height of a triangle **Step 3. Name.**Choose a variable to represent it. Let (h=) the height. Draw the figure and label it with the given information. Area = 126 sq. in. **Step 4. Translate.**Write the appropriate formula. (A=frac{1}{2}bh) Substitute in the given information. (126=frac{1}{2}·18·h) **Step 5. Solve**the equation.(126=9h) Divide both sides by 9. (14=h) **Step 6. Check**.(egin{align*} A &= frac{1}{2}bh 126 & stackrel{?}{=} 12·18·14 126 &=126✓ end{align*})

**Step 7. Answer**the question.The height of the triangle is (14) inches.

EXAMPLE (PageIndex{14}):

The area of a triangular church window is (90) square meters. The base of the window is (15) meters. What is the window’s height?

**Answer**The window’s height is (12) meters.

EXAMPLE (PageIndex{15}):

A triangular tent door has area (15) square feet. The height is five feet. What is the length of the base?

**Answer**The length of the base is (6) feet.

In the next example, we will work with a **right triangle**. To solve for the measure of each angle, we need to use two triangle properties. In any triangle, the sum of the measures of the angles is (180°). We can write this as a formula: (m∠A+m∠B+m∠C=180). Also, since the triangle is a right triangle, we remember that a right triangle has one (90°) angle.

Here, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

EXAMPLE (PageIndex{16}):

The measure of one angle of a right triangle is 40 degrees more than the measure of the smallest angle. Find the measures of all three angles.

**Answer****Step 1. Identify**what you are looking for.the measures of all three angles **Step 3. Name.**Choose a variable to represent it.(egin{align*} ext{Let }a ; & = ; mathrm{1^{st} ; angle.} a+40 &= mathrm{2^{nd} ; angle} 90 &= mathrm{3^{rd} ; angle ; (the ; right ; angle)} end{align*}) Draw the figure and label it with the given information. **Step 4. Translate.**Write the appropriate formula. Substitute into the formula. **Step 5. Solve**the equation.**Step 6. Check**. ( egin{align*} 25+65+90 & stackrel{?}{=} 180 180 &= 180✓ end{align*})**Step 7. Answer**the question.The three angles measure (25°,;65°), and (90°).

EXAMPLE (PageIndex{17}):

The measure of one angle of a right triangle is 50 more than the measure of the smallest angle. Find the measures of all three angles.

**Answer**The measures of the angles are (20°, ;70°), and (90°).

EXAMPLE (PageIndex{18}):

The measure of one angle of a right triangle is (30) more than the measure of the smallest angle. Find the measures of all three angles.

**Answer**The measures of the angles are (30°,; 60°), and (90°).

The next example uses another important geometry formula. The **Pythagorean Theorem** tells how the lengths of the three sides of a right triangle relate to each other. Writing the formula in every exercise and saying it aloud as you write it may help you memorize the Pythagorean Theorem.

THE PYTHAGOREAN THEOREM

In any right triangle, where *a* and *b* are the lengths of the legs, and *c* is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

We will use the Pythagorean Theorem in the next example.

EXAMPLE (PageIndex{19}):

Use the Pythagorean Theorem to find the length of the other leg in

**Answer****Step 1. Identify**what you are looking for.the length of the leg of the triangle **Step 3. Name.**Choose a variable to represent it. Let (a) = the leg of the triangle. Label side *a*.**Step 4. Translate.**Write the appropriate formula. Substitute. (egin{align*}a^2+b^2 &=c ^2 a^2+12^2 &=13^2 end{align*}) **Step 5. Solve**the equation. Use the definition of square root. Simplify.(egin{align*} a^2+144 &= 169 a^2 &= 25 a &= sqrt{25} a&=5 end{align*}) **Step 6. Check.****Step 7. Answer**the question.The length of the leg is (5).

EXAMPLE (PageIndex{20}):

Use the Pythagorean Theorem to find the length of the leg in the figure.

**Answer**The length of the leg is (8).

EXAMPLE (PageIndex{21}):

Use the Pythagorean Theorem to find the length of the leg in the figure.

**Answer**The length of the leg is (12).

The next example is about the **perimeter** of a rectangle. Since the perimeter is just the distance around the rectangle, we find the sum of the lengths of its four sides—the sum of two lengths and two widths. We can write is as (P=2L+2W) where (L) is the length and (W) is the width. To solve the example, we will need to define the length in terms of the width.

EXAMPLE (PageIndex{22}):

The length of a rectangle is six centimeters more than twice the width. The perimeter is (96) centimeters. Find the length and width.

**Answer****Step 1. Identify**what we are looking for.the length and the width **Step 3. Name**. Choose a variable to represent the width. The length is six more than twice the width.Let (w=) width. (2w+6=) length

(P=96) cm**Step 4. Translate.**Write the appropriate formula. Substitute in the given information. **Step 5. Solve the equation.****Step 6.**Check.(egin{align*} P &=2L+2W 96 & stackrel{?}{=}2·34+2·14 96 &=96✓ end{align*})**Step 7. Answer the question.**The length is (34) cm and the width is (14) cm

EXAMPLE (PageIndex{23}):

The length of a rectangle is seven more than twice the width. The perimeter is (110) inches. Find the length and width.

**Answer**The length is (16) inches and the width is (39) inches.

EXAMPLE (PageIndex{24}):

The width of a rectangle is eight yards less than twice the length. The perimeter is (86) yards. Find the length and width.

**Answer**The length is (17) yards and the width is (26) yards.

The next example is about the **perimeter** of a triangle. Since the perimeter is just the distance around the triangle, we find the sum of the lengths of its three sides. We can write this as (P=a+b+c), where (a), (b), and (c) are the lengths of the sides.

EXAMPLE (PageIndex{25}):

One side of a triangle is three inches more than the first side. The third side is two inches more than twice the first. The perimeter is (29) inches. Find the length of the three sides of the triangle.

**Answer****Step 1. Identify**what we are looking for.the lengths of the three sides of a triangle **Step 3. Choose a variable to represent the length of the first side.**( egin{align*} mathrm{Let ;}x ;& mathrm{= ; length ; of ;1^{st} ;side.} x+3 ; &= ; mathrm{length ; of ; 2^{nd} ; side} 2x+2 ; &= ; mathrm{length ; of ;3^{rd} ; side} end{align*})

**Step 4. Translate.**Write the appropriate formula. Substitute in the given information.**Step 5. Solve**the equation.**Step 6. Check.**(egin{align*} 29 & stackrel{?}{=}6+9+14 29 &= 29✓ end{align*})**Step 7. Answer**the question.The lengths of the sides of the triangle are (6), (9), and (14) inches.

EXAMPLE (PageIndex{26}):

One side of a triangle is seven inches more than the first side. The third side is four inches less than three times the first. The perimeter is (28) inches. Find the length of the three sides of the triangle.

**Answer**The lengths of the sides of the triangle are (5), (11) and (12) inches.

EXAMPLE (PageIndex{27}):

One side of a triangle is three feet less than the first side. The third side is five feet less than twice the first. The perimeter is (20) feet. Find the length of the three sides of the triangle.

**Answer**The lengths of the sides of the triangle are (4), (7) and (9) feet.

EXAMPLE (PageIndex{28}):

The perimeter of a rectangular soccer field is (360) feet. The length is (40) feet more than the width. Find the length and width.

**Answer****Step 1. Identify**what we are looking for.the length and width of the soccer field **Step 3. Name.**Choose a variable to represent it. The length is 40 feet more than the width. Draw the figure and label it with the given information.Let *w*= width. (w+40=) length**Step 4. Translate**. Write the appropriate formula and substitute.**Step 5. Solve**the equation.**Step 6. Check.**( egin{align*} P &=2L+2W 360 & stackrel{?}{=} 2(110)+2(70) 360 &=360✓ end{align*})**Step 7. Answer**the question.The length of the soccer field is (110) feet and the width is (70) feet.

EXAMPLE (PageIndex{29}):

The perimeter of a rectangular swimming pool is (200) feet. Find the length and width.

**Answer**The length of the swimming pool is (70) feet and the width is (30) feet.

EXAMPLE (PageIndex{30}):

The length of a rectangular garden is (30) yards more than the width. The perimeter is (300) yards. Find the length and width.

**Answer**The length of the garden is (90) yards and the width is (60) yards.

Applications of these geometric properties can be found in many everyday situations as shown in the next example.

EXAMPLE (PageIndex{31}):

Kelvin is building a gazebo and wants to brace each corner by placing a 10” piece of wood diagonally as shown.

How far from the corner should he fasten the wood if wants the distances from the corner to be equal? Approximate to the nearest tenth of an inch.

**Answer****Step 1. Identify**what we are looking for.the distance from the corner that the bracket should be attached **Step 3. Draw the figure and label it with the given information.**Let (x=) the distance from the corner. **Step 4. Translate.**Write the appropriate formula and substitute.(a^2+b^2=c^2) (x^2+x^2=10^2) **Step 5. Solve**the equation.Isolate the variable.

Use the definition of square root.Simplify. Approximate to the nearest tenth.

( egin{align*} 2x^2 &= 100 x^2 &=50 x &= sqrt{50} x &≈7.1 end{align*})

**Step 6. Check.**( egin{align*} a^2+b^2 &= c^2 (7.1)^2+(7.1)^2 &≈10^2 ; ;;;; ext{Yes.} end{align*})

**Step 7. Answer**the question.Kelvin should fasten each piece of wood approximately 7.1” from the corner.

EXAMPLE (PageIndex{32}):

John puts the base of a (13)-foot ladder five feet from the wall of his house as shown in the figure. How far up the wall does the ladder reach?

**Answer**The ladder reaches (12) feet.

EXAMPLE (PageIndex{33}):

Randy wants to attach a (17)-foot string of lights to the top of the (15) foot mast of his sailboat, as shown in the figure. How far from the base of the mast should he attach the end of the light string?

**Answer**He should attach the lights (8) feet from the base of the mast.

Access this online resource for additional instruction and practice with solving for a variable in literal equations.

- Solving Literal Equations

## Key Concepts

**How To Solve Geometry Applications****Read**the problem and make sure all the words and ideas are understood.**Identify**what you are looking for.**Name**what you are looking for by choosing a variable to represent it. Substitute in the given information.**Solve**the equation using good algebra techniques.**Check**the answer in the problem and make sure it makes sense.**Answer**the question with a complete sentence.

**The Pythagorean Theorem**- In any right triangle, where
*a*and*b*are the lengths of the legs, and*c*is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

- In any right triangle, where

## 5.6: Solve a Formula for a Specific Variable

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## Solve for a Variable in a Formula

In these lessons, we will learn how to solve a formula for a variable. It is also called solving literal equations.

Solving for a variable in a formula involves similar steps to solving for a variable in an equation. Let us review the steps as follows when we need to solve for a variable in a formula.

The formula for the area of a circle is *A* = p*r* 2 . Solve for *r* in the formula .

The formula for the area of a circle is *A* = p*r* 2 . Solve for *r* in the formula .

**How to solve a formula with variables for a specified variable?**

Examples:

1. Solve for r

d = rt

5. Write y in terms of x

y - 3 = 1/3(x - 4)

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Yours is a non linear equation . So you can use optimize.fsolve for it. For further details look for the function in this tutorial scipy

(I don't know why you mention scipy in your question when you use sympy in your code. I'll assume you are using sympy.)

Sympy can solve this equation if you specify an integer power for y (ie y**3.0 changed to y**3 ).

The following works for me using Sympy 0.6.7.

Assuming you mean you were trying to use sympy, as opposed to scipy, then you can get Sympy (works with v0.7.2+) to solve it by making a small adjustment to way you defined your equation - you just need to put a multiplication operator (*) in between the first 'y' and the '('. It doesn't appear to matter whether you specify the power as a float or not (but it's possible it was required in 0.6.7).

## Solving Formulas For A Specific Variable Worksheet / 1

**Solving Formulas For A Specific Variable Worksheet / 1**. We will call this solving an equation for a specific variable in general. (more) is ideal for teachers with one computer in their room. This video shows how to solve a formula with variables for a specified variable. These familiar, frequently used equations are called formulas. A few of the equations might be familiar from the study of linear equations and slope.

We have all probably worked with some geometric formulas in our study of mathematics. Solving for single variable proving to be extremely difficult. We will call this solving an equation for a specific variable in general. This process is also called solving a literal equation. Worksheets are given each formula below solve each equation for the, literal equations and formulas, work 2 2 solving equations in one variable, practice solving literal equations, solving multi step equations.

Rearranging Equations Worksheets Literal Equations from www.mathworksheets4kids.com We have all probably worked with some geometric formulas in our study of mathematics. By the way, concerning formulas solving for variables worksheet, scroll the page to see particular variation of pictures to add more info. Trigonometry solver with showing work, method of substitution calculator, equality of matrices worksheet. Solving for a variable worksheet as a derivative of big methods answer questions. In cases where you actually require help with math and in particular with variable or formulas come pay occasionally it is necessary to take a formula and rearrange it so that it is solved for a different variable. Some of the worksheets for this concept are given each formula below solve each equation for the, literal equations and formulas, work 2 2 solving equations in one variable, practice solving literal equations, solving. The formula contains letters, or literals. I would like to solve this formula for x.

### In cases where you actually require help with math and in particular with variable or formulas come pay occasionally it is necessary to take a formula and rearrange it so that it is solved for a different variable.

Solving formulas for a specific variable. Solving for a specific variable. Solving for single variable proving to be extremely difficult. Just in case you need help on solving systems or scientific, mathscitutor.com happens to be the right destination to head to! (more) is ideal for teachers with one computer in their room. This can be done by multiplying the entire fraction by the least common isolate the variable we are solving for, this is generally done by dividing both sides of the equation. Solving for an equation for one variable (algebra). So, i want to set the last active worksheet as a variable. Formulas are used in so many fields, it is important to recognize if you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. Scholars solve literal equations for specified variables. Practice solving formulas and equations for specific variables. I would like to solve this formula for x. Solving for a variable in a formula involves similar steps to solving for a variable in an equation.

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Solving Linear Equations Part Ii from saylordotorg.github.io So, i want to set the last active worksheet as a variable. This is an extension of the exploration guide with work for students to do at their desk. By the way, concerning formulas solving for variables worksheet, scroll the page to see particular variation of pictures to add more info. Some variables are just more interesting than others. Solving for single variable proving to be extremely difficult. Solving formulas for a variable. This concept requires an understanding of variables and intro level algebra concepts. The reason that i want to do that is because i run this macro as you can see the active worksheet changes two times.

### Worksheet works solving multi step equations variables both from solving for a variable worksheet, source:koogra.com.

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Source: helpingwithmath.com

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Source: ecdn.teacherspayteachers.com

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(more) is ideal for teachers with one computer in their room.

Solving formulas for a variable—task rubric (.doc).

Choose the correct steps to solve a formula for a given variable.

Practice solving formulas and equations for specific variables.

Pocketmath.net provides good tips on solving for a variable worksheet, subtracting rational and multiplying and other math subjects.

How to solve a formula with variables for a specified variable?

Using worksheets suggests facilitating students to manage to solution issues about matters they've learned.

Source: mathskills4kids.com

We will call this solving an equation for a specific variable in general.

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Source: www.mathworksheets4kids.com

These 10 worksheets involve solving the variables.

Source: naacpcharlestonbranch.org

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The reason that i want to do that is because i run this macro as you can see the active worksheet changes two times.

· rearrange formulas to isolate specific variables.

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How to solve a formula with variables for a specified variable?

Source: algebra-equation.com

How to solve a formula with variables for a specified variable?

Source: d2pbkktgz4wpxb.cloudfront.net

Solving for a variable in a formula involves similar steps to solving for a variable in an equation.

Source: helpingwithmath.com

Solving quadratic equations by quadratic formula.

This video shows how to solve a formula with variables for a specified variable.

We will call this solving an equation for a specific variable in general.

This can be done by multiplying the entire fraction by the least common isolate the variable we are solving for, this is generally done by dividing both sides of the equation.

Solving quadratic equations by quadratic formula.

Source: helpingwithmath.com

We will call this solving an equation for a specific variable in general.

Source: naacpcharlestonbranch.org

Since we are solving rational formulas or formulas containing fractions, the first thing we need to do is to get rid of the fractions.

## Solve for Variable Calculator

The variables in the literal equation represents the known values. Solving a given formula for a particular variable is called as literal equations. The solve for variable calculator identifies the variables then add or subtract variables on one side and add or subtract the whole numbers on other side. Finally brings the numbers to one side and multiply or divide to find the value of the variable. Find the variable by solving literal equation using the literal equation calculator.

## A tougher sequence

Let&rsquos try another problem. How many diagonals does a regular n -gon have? (Recall that a regular n -gon is a shape with n equal sides and n equal angles. A is a line from one vertex of the n -gon to another, except that edges of the n -gon don&rsquot count.)

As before, we start by drawing some examples. The smallest value of n that makes sense is n = 3 (a 3-gon is a triangle) we continue up to n = 8 (an 8-gon is an octagon).

We see that a triangle has no diagonals, a square has two, a pentagon has five, a hexagon has nine, and so on. We record this information in a table, using n for the number of sides and d ( n ) for the number of diagonals. [We could have reused f ( n ), but since we already used the letter f with a different meaning earlier we chose a different letter this time. On the other hand, we are reusing the letter n with a different meaning here. It&rsquos not terribly important what you name your variables just be consistent within any one problem.]

As before, we&rsquoll start by taking differences between successive terms of the sequence.

n : | 3 | 4 | 5 | 6 | 7 | 8 | ||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|

d ( n ): | 0 | 2 | 5 | 9 | 14 | 20 | ||||||

2 | 3 | 4 | 5 | 6 |

This time we don&rsquot have constant differences. But we notice that the differences themselves seem to have a pattern&mdasheach difference is 1 more than the previous difference. So, if we take the differences of the differences, we will get a constant row:

n : | 3 | 4 | 5 | 6 | 7 | 8 | ||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|

d ( n ): | 0 | 2 | 5 | 9 | 14 | 20 | ||||||

2 | 3 | 4 | 5 | 6 | ||||||||

1 | 1 | 1 | 1 |

Since we had to take differences *twice* before we found a constant row, we guess that the formula for the sequence is a polynomial of degree 2 , i.e., a polynomial. (In general, if you have to take differences m times to get a constant row, the formula is probably a polynomial of degree m .) The general form of a function given by a quadratic polynomial is

where the coefficients a , b , and c are real numbers. In our case these coefficients are unknown we are trying to find them. Since we have three unknowns, we will need a system of three equations. To get these equations, we can use some of the values of n and d ( n ) that we know. For example, plugging in n = 3, d ( n ) = 0 into the general quadratic d ( n ) = a n 2 + b n + c gives the equation

Similarly, plugging in n = 4, d ( n ) = 2 gives

This gives us the following simultaneous system of three linear equations in three unknowns. (The numbers in parentheses to the left of the equations are simply labels so that we can refer to the equations they are not part of the equations themselves.)

(1) | 9 a | + | 3 b | + | c | = | 0 |

(2) | 16 a | + | 4 b | + | c | = | 2 |

(3) | 25 a | + | 5 b | + | c | = | 5 |

We can solve this system algebraically to find the values of the coefficients a , b , and c . (For a review of some useful algebraic techniques, see Appendix A in Problem Solving Through Recreational Mathematics .)

One way to solve a system of linear equations is to systematically substitute or eliminate variables one at a time until only one variable remains. Solving for the single remaining variable is easy, and then you can back-substitute the values of the variables you know to find the values of the variables that were eliminated.

Let&rsquos start by solving equation (1) for c .

This gives us an expression to substitute for the variable c . If we plug this expression into the other two equations, (2) and (3) , we get a simultaneous system of two linear equations in two unknowns.

16 a | + | 4 b | + | (&minus9 a &minus 3 b ) | = | 2 |

25 a | + | 5 b | + | (&minus9 a &minus 3 b ) | = | 5 |

After we combine like terms, we get the following system.

Now b looks like a good candidate to eliminate. If we multiply equation (5) by &minus2 and add it to equation (6) , the b &rsquos will cancel:

&minus14 a | &minus | 2 b | = | &minus4 |

16 a | + | 2 b | = | 5 |

(7) | 2 a | = | 1 |

We have substituted or eliminated all but one variable, whose value we can now find. From equation (7) we see that a = 0.5. Now we begin the process of back-substituting the values of variables we know into previous equations in order to find the values of the other variables. We can plug the value a = 0.5 into equation (6) , say, to get the equation

so 2 b = 5 &minus 8 = &minus3, which means b = &minus1.5. Now we know the values of both a and b , so we can plug them into equation (4) to get

So we have found the coefficients of the quadratic polynomial: a = 0.5, b = &minus1.5, and c = 0. So we guess that the formula for d ( n ) is

Let&rsquos check this formula with some data from our table, to make sure we didn&rsquot make an algebraic error somewhere. We&rsquoll try n = 8, for which we should get d ( n ) = 20:

It checks out. Try a few more values of n and make sure this formula works for the data we have.

## Count cells that contain specific text (case-sensitive)

In situation when you need to differentiate uppercase and lowercase characters, the COUNTIF function won't work. Depending on whether you are looking for an exact or partial match, you will have to build a different formula.

### Case-sensitive formula to count cells with specific text (exact match)

To count the number of cells with certain text recognizing the text case, we will use a combination of the SUMPRODUCT and EXACT functions:

- EXACT compares each cell in the range against the sample text and returns an array of TRUE and FALSE values, TRUE representing exact matches and FALSE all other cells. A double hyphen (called a
*double unary*) coerces TRUE and FALSE into 1's and 0's. sums all the elements of the array. That sum is the number of 1's, which is the number of matches.

For example, to get the number of cells in A2:A10 that contain the text in D1 and handle uppercase and lowercase as different characters, use this formula:

### Case-sensitive formula to count cells with specific text (partial match)

To build a case-sensitive formula that can find a text string of interest anywhere in a cell, we are using 3 different functions:

- The case-sensitive FIND function searches for the target text in each cell of the range. If it succeeds, the function returns the position of the first character, otherwise the #VALUE! error. For the sake of clarity, we do not need to know the exact position, any number (as opposed to error) means that the cell contains the target text.
- The ISNUMBER function handles the array of numbers and errors returned by FIND and converts the numbers to TRUE and anything else to FALSE. A double unary (--) coerces the logical values into ones and zeros.
- SUMPRODUCT sums the array of 1's and 0's and returns the count of cells that contain the specified text as part of their contents.

To test the formula on real-life data, let's find how many cells in A2:A10 contain the substring input in D1:

And this returns a count of 3 (cells A2, A3 and A6):

## MathHelp.com

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:

#### Solve A = bh for b

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 = 2*b* 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.

#### Solve d = rt for r

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* :

You should always check that your "solution" really **is** a solution.

### How To Check

Take the solution(s) and put them in the **original equation** to see if they really work.

### Example: solve for x:

*2x***x &minus 3** + 3 = *6***x &minus 3** (x&ne3)

We have said x&ne3 to avoid a division by zero.

Let's multiply through by (x &minus 3) :

That can be solved by having x=3

*2 × 3* **3 &minus 3** + 3 = *6* **3 &minus 3**

**Hang On! That means Dividing by Zero!**

And anyway, we said at the top that x&ne3 , so .

x = 3 does not actually work, and so:

There is **No** Solution!

That was interesting . we **thought** we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!