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Calculator Use

Solve math problems using order of operations like PEMDAS, BEDMAS, BODMAS, GEMDAS and MDAS. ( PEMDAS Caution ) This calculator solves math equations that add, subtract, multiply and divide positive and negative numbers and exponential numbers. You can also include parentheses and numbers with exponents or roots in your equations.

Use these math symbols:

+  Addition -  Subtraction *  Multiplication /  Division ^  Exponents (2^5 is 2 raised to the power of 5) r  Roots (2r3 is the 3rd root of 2) () [] {}  Brackets or Grouping

You can try to copy equations from other printed sources and paste them here and, if they use ÷ for division and × for multiplication, this equation calculator will try to convert them to / and * respectively but in some cases you may need to retype copied and pasted symbols or even full equations.

If your equation has fractional exponents or roots be sure to enclose the fractions in parentheses. For example:

Entering fractions

If you want an entry such as 1/2 to be treated as a fraction then enter it as (1/2). For example, in the equation 4 divided by ½ you must enter it as 4/(1/2). Then the division 1/2 = 0.5 is performed first and 4/0.5 = 8 is performed last. If you incorrectly enter it as 4/1/2 then it is solved 4/1 = 4 first then 4/2 = 2 last. 2 is a wrong answer. 8 was the correct answer.

Math Order of Operations - PEMDAS, BEDMAS, BODMAS, GEMDAS, MDAS

PEMDAS is an acronym that may help you remember order of operations for solving math equations. PEMDAS is typcially expanded into the phrase, "Please Excuse My Dear Aunt Sally." The first letter of each word in the phrase creates the PEMDAS acronym. Solve math problems with the standard mathematical order of operations, working left to right:

PEMDAS Caution

Multiplication DOES NOT always get performed before Division. Multiplication and Division are performed as they occur in the equation, from left to right.

Addition DOES NOT always get performed before Subtraction. Addition and Subtraction are performed as they occur in the equation, from left to right.

The order "MD" (DM in BEDMAS) is sometimes confused to mean that Multiplication happens before Division (or vice versa). However, multiplication and division have the same precedence. In other words, multiplication and division are performed during the same step from left to right. For example, 4/2*2 = 4 and 4/2*2 does not equal 1.

The same confusion can also happen with "AS" however, addition and subtraction also have the same precedence and are performed during the same step from left to right. For example, 5 - 3 + 2 = 4 and 5 - 3 + 2 does not equal 0.

A way to remember this could be to write PEMDAS as PE(MD)(AS) or BEDMAS as BE(DM)(AS).

Order of Operations Acronyms

The acronyms for order of operations mean you should solve equations in this order always working left to right in your equation.

PEMDAS stands for " P arentheses, E xponents, M ultiplication and D ivision, A ddition and S ubtraction"

You may also see BEDMAS, BODMAS, and GEMDAS as order of operations acronyms. In these acronyms, "brackets" are the same as parentheses, and "order" is the same as exponents. For GEMDAS, "grouping" is like parentheses or brackets.

BEDMAS stands for " B rackets, E xponents, D ivision and M ultiplication, A ddition and S ubtraction"

BEDMAS is similar to BODMAS.

BODMAS stands for " B rackets, O rder, D ivision and M ultiplication, A ddition and S ubtraction"

GEMDAS stands for " G rouping, E xponents, D ivision and M ultiplication, A ddition and S ubtraction"

MDAS is a subset of the acronyms above. It stands for " M ultiplication, and D ivision, A ddition and S ubtraction"

Operator Associativity

Multiplication, division, addition and subtraction are left-associative. This means that when you are solving multiplication and division expressions you proceed from the left side of your equation to the right. Similarly, when you are solving addition and subtraction expressions you proceed from left to right.

Examples of left-associativity:

Exponents and roots or radicals are right-associative and are solved from right to left.

Examples of right-associativity:

For nested parentheses or brackets, solve the innermost parentheses or bracket expressions first and work toward the outermost parentheses. For each expression within parentheses, follow the rest of the PEMDAS order: First calculate exponents and radicals, then multiplication and division, and finally addition and subtraction.

You can solve multiplication and division during the same step in the math problem: after solving for parentheses, exponents and radicals and before adding and subtracting. Proceed from left to right for multiplication and division. Solve addition and subtraction last after parentheses, exponents, roots and multiplying/dividing. Again, proceed from left to right for adding and subtracting.

Adding, Subtracting, Multiplying and Dividing Positive and Negative Numbers

This calculator follows standard rules to solve equations.

Rules for Addition Operations (+)

If signs are the same then keep the sign and add the numbers.

If signs are different then subtract the smaller number from the larger number and keep the sign of the larger number.

Rules for Subtraction Operations (-)

Keep the sign of the first number. Change all the following subtraction signs to addition signs. Change the sign of each number that follows so that positive becomes negative, and negative becomes positive then follow the rules for addition problems.

Rules for Multiplication Operations (* or ×)

Multiplying a negative by a negative or a positive by a positive produces a positive result. Multiplying a positive by a negative or a negative by a positive produces a negative result.

Rules for Division Operations (/ or ÷)

Similar to multiplication, dividing a negative by a negative or a positive by a positive produces a positive result. Dividing a positive by a negative or a negative by a positive produces a negative result.

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Furey, Edward " Math Equation Solver | Order of Operations " at https://www.calculatorsoup.com/calculators/math/math-equation-solver.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators

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10 Math Equations That Have Never Been Solved

By Kathleen Cantor, 10 Sep 2020

Mathematics has played a major role in so many life-altering inventions and theories. But there are still some math equations that have managed to elude even the greatest minds, like Einstein and Hawkins. Other equations, however, are simply too large to compute. So for whatever reason, these puzzling problems have never been solved. But what are they?

Like the rest of us, you're probably expecting some next-level difficulty in these mathematical problems. Surprisingly, that is not the case. Some of these equations are even based on elementary school concepts and are easily understandable - just unsolvable.

1. The Riemann Hypothesis

Equation: σ (n) ≤ Hn +ln (Hn)eHn

For an instance, if n = 4 then σ(4)=1+2+4=7 and H4 = 1+1/2+1/3+1/4. Solve this equation to either prove or disprove the following inequality n≥1? Does it hold for all n≥1?

This problem is referred to as Lagarias’s Elementary Version of the Riemann Hypothesis and has a price of a million dollars offered by the  Clay Mathematics Foundation  for its solution.

2. The Collatz Conjecture

Equation: 3n+1

Prove the answer end by cycling through 1,4,2,1,4,2,1,… if n is a positive integer. This is a repetitive process and you will repeat it with the new value of n you get. If your first n = 1 then your subsequent answers will be 1, 4, 2, 1, 4, 2, 1, 4… infinitely. And if n = 5 the answers will be 5,16,8,4,2,1 the rest will be another loop of the values 1, 4, and 2.

This equation was formed in 1937 by a man named Lothar Collatz which is why it is referred to as the Collatz Conjecture.

3. The Erdős-Strauss Conjecture

Equation: 4/n=1/a+1/b+1/c

This equation aims to see if we can prove that for if n is greater than or equal to 2, then one can write 4*n as a sum of three positive unit fractions.

This equation was formed in 1948 by two men named Paul Erdős and Ernst Strauss which is why it is referred to as the Erdős-Strauss Conjecture.

4. Equation Four

Equation: Use 2(2∧127)-1 – 1 to prove or disprove if it’s a prime number or not?

Looks pretty straight forward, does it? Here is a little context on the problem.

Let’s take a prime number 2. Now, 22 – 1 = 3 which is also a prime number. 25 – 1 = 31 which is also a prime number and so is 27−1=127. 2127 −1=170141183460469231731687303715884105727 is also prime.

5. Goldbach's Conjecture

Equation: Prove that x + y = n

This problem, as relatively simple as it sounds has never been solved. Solving this problem will earn you a free million dollars. This equation was first proposed by Goldbach hence the name Goldbach's Conjecture.

If you are still unsure then pick any even number like 6, it can also be expressed as 1 + 5, which is two primes. The same goes for 10 and 26.

6. Equation Six

Equation: Prove that (K)n = JK1N(q)JO1N(q)

This equation tries to portray the relationship between  quantum invariants  of knots and  the hyperbolic geometry  of  knot complements . Although this equation is in mathematics, you have to be a physics familiar to grasp the concept.

7. The Whitehead Conjecture

Equation: G = (S | R)

What you are doing in this equation is prove the claim made by Mr.  Whitehead  in 1941 in  an algebraic topology  that every subcomplex of an  aspherical   CW complex  that is connected and in two dimensions is also spherical. This was named after the man, Whitehead conjecture.

8. Equation Eight

Equation: (EQ4)

This equation is the definition of  morphism  and is referred to as an assembly map.  Check out the  reduced C*-algebra  for more insight into the concept surrounding this equation.

9. The Euler-Mascheroni Constant

Equation: y=limn→∞(∑m=1n1m−log(n))

Find out if y is rational or irrational in the equation above. To fully understand this problem you need to take another look at rational numbers and their concepts.  The character y is what is known as the Euler-Mascheroni constant and it has a value of 0.5772.

This equation has been calculated up to almost half of a trillion digits and yet no one has been able to tell if it is a rational number or not.

10. Equation Ten

Equation: π + e

Find the sum and determine if it is algebraic or transcendental. To understand this question you need to have an idea of  algebraic real numbers  and how they operate. The number pi or π originated in the 17th century and it is transcendental along with e. but what about their sum? So Far this has never been solved.

As you can see in the equations above, there are several seemingly simple mathematical equations and theories that have never been put to rest. Decades are passing while these problems remain unsolved. If you're looking for a brain teaser, finding the solutions to these problems will give you a run for your money.

See the 11 Comments below.

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Posted in Mathematics category - 10 Sep 2020 [ Permalink ]

11 Comments on “10 Math Equations That Have Never Been Solved”

But 2(2127)−1 = 340282366920938463463374607431768211455 is not a prime number. It is divisible by 64511.

Hello I am explorer and i type on google search " unsolvable mathematical formulas ", and I first find this syte. I see you are good-math-guys. Do you know what is this formula means:

π × ∞ = " 5 "

If you happen to have a quantum computer, I am not kidding be smart and don't insert this formula: [π × ∞ = " 5 "] into it please.

Maybe only, if you know meaning of this three symbols up writen and connected together.

(x dot epsilon)

I can explain my theory if you want me to spoil the pleasure of solving the equation. And mathematics as a science too or " as well " sorry i am not good in English, and google translate is not exelent.

8.539728478 is the answer to number 10

8.539728478 is the answer to number 10 or 8.539734221

Equation Four: Solved

To determine whether the number 2(2^127)-1 – 1 is a prime number, we first need to calculate its value. The expression 2(2^127) can be simplified as follows:

2(2^127) = 2 * 2^127 = 2^128

Therefore, the expression 2(2^127)-1 – 1 can be written as 2^128 – 1 – 1. We can then simplify this further to get:

2^128 – 1 – 1 = 2^128 – 2

To determine whether this number is prime, we can use the fundamental theorem of arithmetic, which states that every positive integer can be written as a product of prime numbers in a unique way (ignoring the order of the factors). This means that if a number is not prime, it can be expressed as the product of two or more prime numbers.

We can use this theorem to determine whether 2^128 – 2 is prime by trying to express it as the product of two or more prime numbers. However, it is not possible to do this, because 2^128 – 2 cannot be evenly divided by any prime number (except for 1, which is not considered a prime number).

Therefore, we can conclude that 2^128 – 2 is a prime number, because it cannot be expressed as the product of two or more prime numbers.

Equation Ten: Solved

The sum of π and e is equal to π + e = 3.14159 + 2.71828 = 5.85987.

To determine whether this number is algebraic or transcendental, we first need to understand the difference between these two types of numbers. Algebraic numbers are numbers that can be expressed as a root of a polynomial equation with integer coefficients, while transcendental numbers cannot be expressed in this way.

In this case, the number 5.85987 can be expressed as the root of the polynomial equation x^2 - 5.85987x + 2.71828 = 0. Therefore, it is an algebraic number.

In conclusion, the sum of π and e is equal to 5.85987, which is an algebraic number.

Equation 2: SOLVED

The equation 3n + 1 states that a positive integer n should be multiplied by 3 and then 1 should be added to the result. If the resulting value is then divided by 2 and the quotient is a non-negative integer, the process should be repeated with the new value of n.

To prove that this equation always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can start by substituting a value for n and performing the calculations as specified in the equation.

For example, if n is equal to 1, the sequence of values will be: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2 = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5

Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5

The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5

The value of n becomes 5 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 5 3n + 1 = 3(5) + 1 = 16 n = 16/2 = 8 n = 8/2 = 4 n = 4/2 = 2 n = 2/2 = 1 n = 1/2 = 0.5

Since n must be a non-negative integer, the value of n becomes 1 when the result of the previous calculation is divided by 2. The sequence then becomes: n = 1 3n + 1 = 3(1) + 1 = 4 n = 4/2

To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.

If n is equal to 4, the sequence of values will be: n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5

Since n must be a non-negative integer, the value of n becomes 6 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 6 3n + 1 = 3(6) + 1 = 19 n = 19/2 = 9.5

Certainly! To complete the proof that the equation 3n + 1 always results in a repeating sequence of 1, 4, 2, 1, 4, 2, 1, ... if n is a positive integer, we can continue the process of substituting values for n and performing the calculations as specified in the equation.

If n is equal to 4, the sequence of values will be:

n = 4 3n + 1 = 3(4) + 1 = 13 n = 13/2 = 6.5

Since n must be a non-negative integer, the value of n becomes 9 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 9 3n + 1 = 3(9) + 1 = 28 n = 28/2 = 14 n = 14/2 = 7 n = 7/2 = 3.5

The value of n becomes 3 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 3 3n + 1 = 3(3) + 1 = 10 n = 10/2 = 5 n = 5/2 = 2.5

Since n must be a non-negative integer, the value of n becomes 2 when the result of the previous calculation is divided by 2. The sequence then becomes:

n = 2 3n + 1 = 3(2) + 1 = 7 n = 7/2 = 3.5

As we can see, the sequence of values becomes repetitive

The Riemann Hypothesis

This equation states that the sum of the positive integers divisible by n (σ(n)) is less than or equal to the n-th harmonic number (Hn) plus the natural logarithm of the n-th harmonic number (ln(Hn)) multiplied by the n-th harmonic number (Hn) raised to the power of Hn.

To solve this equation, you would need to substitute a specific value for n and determine the value of Hn and σ(n) for that specific value. You can then substitute these values into the equation and see if it holds true.

For example, if n = 5, the sum of the positive integers divisible by 5 (σ(5)) is 15 (1 + 5 + 10 + 15 + 20 + 25), the 5th harmonic number (H5) is 2.28, and the natural logarithm of the 5th harmonic number (ln(H5)) is 0.83. Substituting these values into the equation, we get:

σ(5) ≤ H5 + ln(H5)eH5 15 ≤ 2.28 + 0.83 * 2.28^2.28 15 ≤ 4.39

Since 15 is less than or equal to 4.39, the equation holds true for this specific value of n.

Equation #9

In the equation y = limn→∞(∑m=1n1m−log(n)), y is the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity.

The Euler-Mascheroni constant is defined as the limit of the sequence (∑m=1n1m−log(n)) as n approaches infinity, and it has a value of approximately 0.5772. Therefore, y is equal to the Euler-Mascheroni constant, which is a rational number.

Rational numbers are numbers that can be expressed as the ratio of two integers, such as 3/4, 7/11, or 2/5. They can be written as a finite or repeating decimal, such as 0.75, 0.636363636..., or 1.5.

Irrational numbers are numbers that cannot be expressed as the ratio of two integers, and they cannot be written as a finite or repeating decimal. Examples of irrational numbers include √2, π, and e.

Since y is equal to the Euler-Mascheroni constant, which is a rational number, y is a rational number.

The equation G = (S | R) is a definition of a CW complex, where S and R are subcomplexes of G. A CW complex is a topological space that can be built up from cells, where each cell is homeomorphic to a closed ball in Euclidean space.

The statement "when CW complex K (S | R) is aspherical" means that the complex K (S | R) does not contain any non-trivial loops, i.e. loops that cannot be continuously contracted to a point. This implies that the fundamental group of K (S | R) is trivial, which means that π1(K (S | R)) = {e}.

The statement "if π2 (K (S | R)) = 0" means that the second homotopy group of the complex K (S | R) is trivial, which means that there are no non-trivial 2-dimensional holes in K (S | R).

Together, these statements imply that the CW complex K (S | R) is a topological space with no non-trivial loops or holes. This is a strong condition that is satisfied by very few spaces, and it is a necessary condition for a space to be aspherical.

In summary, the statement "when CW complex K (S | R) is aspherical" and "if π2 (K (S | R)) = 0" implies that the complex K (S | R) is a topological space with no non-trivial loops or holes, which is a necessary condition for a space to be aspherical.

#3 Erdos Strauss Conjecture:

To solve the equation 4/n = 1/a + 1/b + 1/c where n ≥ 2, a, b and c are positive integers, we can first multiply both sides of the equation by nabc to get rid of the fractions:

4abc = nab + nbc + nac

We can then group like terms:

4abc = (n + a)(b + c)

Now we can use the fact that n, a, b, and c are positive integers to make some observations:

Since n, a, b and c are positive integers, n, a, b and c must be factors of 4abc. Since n is greater than or equal to 2, it must be one of the factors of 4abc. The other factors of 4abc are (n + a), b, and c. So, to find all the possible values of n, a, b, and c, we must find all the ways to factorize 4abc such that one of the factors is greater than or equal to 2.

4abc = 4 * 1 * 1 * 2 * 3 * 5 = 120

Some possible factorizations are:

n = 2, a = 1, b = 5, c = 12 n = 2, a = 3, b = 5, c = 8 n = 2, a = 4, b = 3, c = 15 n = 2, a = 6, b = 2, c = 20 n = 4, a = 1, b = 3, c = 30 So, the possible solutions to the equation are: (n,a,b,c) = (2,1,5,12), (2,3,5,8), (2,4,3,15), (2,6,2,20), (4,1,3,30)

It's worth noting that this is not an exhaustive list, but just some of the possible solutions, as there could be infinitely many solutions to this equation.

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What is an equation.

An equation is a mathematical statement that describes a relationship between two or more quantities, usually in terms of equalities.

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Example of an Equations

Solving equations is a fundamental part of mathematics.

In mathematics, an equation is a statement that two mathematical expressions are equal to each other. In algebra, the equation usually takes the form of “equals” or “x=”.

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Equations are mathematical expressions that contain an equal sign (=). They are used to show that two expressions are the same.

There are many types of equations and they can be classified according to their operations. For example, there are linear equations, quadratic equations, exponential equations and logarithmic equations.

Linear Equations: Linear Equations can be solved by using the slope-intercept form of the equation. This type of equation is also called a straight line equation because it is represented by a straight line on a graph. It has an equation in two variables, y and x, with an equal sign (=) between them. The slope-intercept form starts with y=mx+b where b is the y-intercept and m is the slope of the line. The slope intercept form can be read as “y is proportional to x plus b” or “y equals mx plus b” or “y varies directly as

Piecewise Functions and How They Can Help With Solving Different Types of Equations

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The piecewise function is one of the most powerful tools in the math tool belt, and can be used to solve a wide variety of equations. In this article, we will explore how to use a piecewise function to solve for a linear equation.

Types of Questions Asked on the ACT & SAT Math Subject Tests (keyword topics for this blog post include quadratic equations, coordinate plane graphs etc.)

The ACT and SAT math subject tests are both standardized tests that measure knowledge in mathematics. The ACT math test is typically taken by high school juniors and seniors, while the SAT math subject test is typically taken by high school sophomores, juniors, and seniors. The questions on these tests are designed to be challenging for students who are studying mathematics in college or graduate school.

Questions on the ACT Math Subject Test:

– What is the area of a triangle with base 3 and height 5?

– What is the sum of two numbers whose product is -312?

– If x = -3/5, what does y equal?

Questions on the SAT Math Subject Test:

– What number should replace this “?” in order to make this equation true? ?x + 4 = 12 ?

– Find an equation for a line that passes through point (2,-1) and has slope -1/2.

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solve math problems equations

Solve math equations with Math Assistant in OneNote

Write or type any math problem and Math Assistant in OneNote can solve it for you—helping you reach the solution quickly, or displaying step-by-step instructions that help you learn how to reach the solution on your own. After solving your equation, there are many options to continue exploring math learning with Math Assistant.

Note:  OneNote Desktop and OneNote for iPad have a new look! Make sure you've selected the tab with instructions for the for the OneNote version you are using. Equation solving is only available if you have a  Microsoft 365 subscription . If you are a Microsoft 365 subscriber, make sure you have the latest version of Office .

Step 1: Enter your equation

In the  Draw  tab, write or type your equation. Use the  Lasso Select  tool to draw a circle around the equation. Then select  Math . This will open the Math Assistant pane.

Learn more:   Create your equation using ink or text.

Type in the number of questions for the practice quiz.

Step 2: Solve your equation

To solve the current equation, do any of the following:

Click or tap the  Select an action  box and then choose the action you want Math Assistant to take. The available choices in this drop-down menu depend on the selected equation.

Learn more: check the Supported Equations tab of this page. 

Review the solution that OneNote displays underneath the action you selected. In the example below, the selected option  Solve for x  displays the solution.

The Show Steps button in the Math task pane

To learn how OneNote solved the problem, you can click or tap Show steps , and then select the detail of what you want to view. The available choices in this drop-down menu depend on the selected equation.

Icon for Immersive Reader

Generate a practice quiz to keep practicing this type of equation.

Warning:  Generate practice quiz  is not currently available as we are working to optimize the experience. The ability to generate practice quizzes will return later this year. 

Solution steps in the Math Assistant task pane

Tip:  You can drag the solution steps to any place on your page.

Create math equations using ink or text with Math Assistant in OneNote

Problem types supported by Math Assistant

Draw graphs of math functions with Math Assistant in OneNote

Note:  This feature is only available if you have a  Microsoft 365 subscription  for enterprise or education. If you are a Microsoft 365 subscriber, make sure you have the latest version of Office .

In the  Draw  tab, write or type your equation. Use the  Lasso Select  tool to draw a circle around the equation.

screenshot showing the use of the lasso select tool to circle a handwritten equation

Next, from the Draw tab, select  Math . This will open the Math Assistant pane.  

Learn more:

Create your equation using ink or text.

Write an equation or formula

screenshot of the math pane of onenote desktop with the equation (x+3)^2=4. In the Select an action section, the only action available is Solve fore x.

Based on your equation, options for actions will be provided. Select your desired action. 

Your equation and the solution will be displayed in the Math pane. 

Tip:  Select Insert math on page to transfer your results to the OneNote page you are working on. 

Learn more: Check the Supported Equations tab of this page. 

screenshot of the math panel of OneNote desktop. the solution to the equation (x+3)^2=4 is shown. Options for viewing the steps to solve are provided including steps using factoring, factoring by grouping, the quadratic formula, and completing the square.

Step 3: Learn from Math Assistant

To learn how OneNote solved the problem, select the method you'd like to learn about from the provided options. 

Steps for various methods are provided based on your equation.  

Screenshot of the math pane in OneNote desktop. Solution steps are shown for using the completing the square method to solve for x.

Turn Math Assistant on or off in OneNote Class Notebook

Problem Types Supported by Math Assistant

When you use Math Assistant in OneNote , you'll notice that the Select an action dropdown beneath the equation changes depending on your selected equation. Here are some of the problem types supported depending on the equation you're trying to solve.

Note:  This feature is only available if you have a Microsoft 365 subscription . If you are a Microsoft 365 subscriber, make sure you have the latest version of Office .

Create a Math quiz in Microsoft Forms

Generate a practice math quiz with Math Assistant in OneNote

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solve math problems equations

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solve math problems equations

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur..

Just sign up for free and you're in.

Short Answer

In problems 33-40, solve the equation given in problem 1. 2 t x   d x + ( t 2 - x 2 ) d t = 0.

x 2 + t 2 - C t = 0 and t ≡ 0

Step by Step Solution

Solving the given equation.

The given equation is,

2 t x dx + t 2 - x 2 dt = 0

Simplify the above equation,

dx dt = - t 2 - x 2 2 tx dx dt = x 2 t - t 2 x

localid="1663928794004" v = x t x = vt dx dt = v + t dv dt

Substitute x t = v the dx dt = v + t dv dt and in the equation (1),

v + t dv dt = v 2 - 1 2 v

t dv dt = v 2 - 1 2 v - v t dv dt = v 2 - 1 - 2 v 2 2 v t dv dt = - 1 - v 2 2 v t dv dt = - 1 + v 2 2 v

Cross multiplication on both sides in the above equation,

t dv dt = - 1 + v 2 2 v 2 v 1 + v 2 dv = - dt t

Integrating

Taking integration both sides in the above equation

∫ 2 v 1 + v 2 dv = - ∫ dt t

Solve the above equation,

ln 1 + v 2 = - ln t + ln C ln 1 + v 2 + ln t = ln C 1 + v 2 t = C

Finding the solution

Substitute the v = x t in the above equation,

1 + x t 2 t = C

t 2 + x 2 t t 2 = C x 2 + t 2 = C t x 2 + t 2 - C t = 0

Also note that is t=0 a solution

Hence the solution x 2 + t 2 - C t = 0 is and t ≡ 0

Most popular questions for Math Textbooks

x ≡ 0 In problems 33-40, Solve the equation given in Problem 2.

y - 4 x - 1 2 dx - dy = 0

In problems identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form .

Question: Show that equation (13) reduces to an equation of the form d y d x = G ( a x + b y ) ,

when [Hint: If a 1 b 2 = a 2 b 1 , then a 2 / a 1 = b 2 / b 1 = k , so that a 2 = k a 1 and b 2 = k b 1 .]

In problem 7-16 , solve the equation. dx dt = 3 xt 2

Question: In Problems 1-30, solve the equation.

dy dx = x - y - 1 x + y + 5

Want to see more solutions like these?

Recommended explanations on math textbooks, decision maths, probability and statistics, mechanics maths.

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The algebra section of QuickMath allows you to manipulate mathematical expressions in all sorts of useful ways. At the moment, QuickMath can expand, factor or simplify virtually any expression, cancel common factors within fractions, split fractions up into smaller ('partial') fractions and join two or more fractions together into a single fraction. More specialized commands are on the way.

What is algebra?

Algebra is the branch of elementary mathematics which uses symbols to stand for unknown quantities. In a more basic sense, it consists of solving equations or manipulating expressions which contain symbols (usually letters, like x, y or z) as well as numbers and functions. Although solving equations is really a part of algebra, it is such a big area that it has its own section in QuickMath.

This part of QuickMath deals only with algebraic expressions. These are mathematical statements which contain letters, numbers and functions, but no equals signs. Here are a few examples of simple algebraic expressions :

The expand command is used mainly to rewrite polynomials with all brackets and whole number powers multiplied out and all like terms collected together. In the advanced section, you also have the option of expanding trigonometric functions, expanding modulo any integer and leaving certain parts of the expression untouched whilst expanding the rest.

Go to the Expand page

The factor command will try to rewrite an expression as a product of smaller expressions. It takes care of such things as taking out common factors, factoring by pairs, quadratic trinomials, differences of two squares, sums and differences of two cubes, and a whole lot more. The advanced section includes options for factoring trigonometric functions, factoring modulo any integer, factoring over the field of Gaussian integers (just the thing for those tricky sums of squares), and even extending the field over which factoring occurs with your own custom extensions.

Go to the Factor page

Simplifying is perhaps the most difficult of all the commands to describe. The way simplification is performed in QuickMath involves looking at many different combinations of transformations of an expression and choosing the one which has the smallest number of parts. Amongst other things, the Simplify command will take care of canceling common factors from the top and bottom of a fraction and collecting like terms. The advanced options allow you to simplify trigonometric functions or to instruct QuickMath to try harder to find a simplified expression.

Go to the Simplify page

The cancel command allows you to cancel out common factors in the denominator and numerator of any fraction appearing in an expression. This command works by canceling the greatest common divisor of the denominator and numerator.

Go to the Cancel page

Partial Fractions

The partial fractions command allows you to split a rational function into a sum or difference of fractions. A rational function is simply a quotient of two polynomials. Any rational function can be written as a sum of fractions, where the denominators of the fractions are powers of the factors of the denominator of the original expression. This command is especially useful if you need to integrate a rational function. By splitting it into partial fractions first, the integration can often be made much simpler.

Go to the Partial Fractions page

Join Fractions

The join fractions command essentially does the reverse of the partial fractions command. It will rewrite a number of fractions which are added or subtracted as a single fraction. The denominator of this single fraction will usually be the lowest common multiple of the denominators of all the fractions being added or subtracted. Any common factors in the numerator and denominator of the answer will automatically be cancelled out.

Go to the Join Fractions page

Introduction to Algebraic Functions

The notion of correspondence is encountered frequently in everyday life. For example, to each book in a library there corresponds the number of pages in the book. As another example, to each human being there corresponds a birth date. To cite a third example, if the temperature of the air is recorded throughout a day, then at each instant of time there is a corresponding temperature.

The examples of correspondences we have given involve two sets X and Y. In our first example, X denotes the set of books in a library and Y the set of positive integers. For each book x in X there corresponds a positive integer y, namely the number of pages in the book. In the second example, if we let X denote the set of all human beings and Y the set of all possible dates, then to each person x in X there corresponds a birth date y.

We sometimes represent correspondences by diagrams of the type shown in Figure 1.17, where the sets X and Y are represented by points within regions in a plane. The curved arrow indicates that the element y of Y corresponds to the element x of X. We have pictured X and Y as different sets. However, X and Y may have elements in common. As a matter of fact, we often have X = Y.

solve math problems equations

A function f from a set X to a set Y is a correspondence that assigns to each element x of X a unique element y of Y. The element y is called the image of x under f and is denoted by f(x). The set X is called the domain of the function. The range of the function consists of all images of elements of X.

Earlier, we introduced the notation f(x) for the element of Y which corresponds to x. This is usually read "f of x." We also call f(x) the value of f at x. In terms of the pictorial representation given earlier, we may now sketch a diagram as in Figure 1.18. The curved arrows indicate that the elements f(x), f(w), f(z), and f(a) of Y correspond to the elements x, y, z and a of X. Let us repeat the important fact that to each x in X there is assigned precisely one image f(x) in Y; however, different elements of X such as w and z in Figure 1.18 may have the same image in Y.

solve math problems equations

Solution As in Example 1, finding images under f is simply a matter of substituting the appropriate number for x in the expression for f(x). Thus:

solve math problems equations

Many formulas which occur in mathematics and the sciences determine functions. As an illustration, the formula A = pi*r 2 for the area A of a circle of radius r assigns to each positive real number r a unique value of A. This determines a function f, where f(r) = pi*r 2 , and we may write A= f(r). The letter r, which represents an arbitrary number from the domain off, is often called an independent variable. The letter A, which represents a number from the range off, is called a dependent variable, since its value depends on the number assigned tor. When two variables r and A are related in this manner, it is customary to use the phrase A is a function of r. To cite another example, if an automobile travels at a uniform rate of 50 miles per hour, then the distance d (miles) traveled in time t (hours) is given by d = 50t and hence the distance d is a function of time t.

We have seen that different elements in the domain of a function may have the same image. If images are always different, then, as in the next definition, the function is called one-to-one.  

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  20. Solve math equations with Math Assistant in OneNote

    Step 1: Enter your equation In the Draw tab, write or type your equation. Use the Lasso Select tool to draw a circle around the equation. Then select Math. This will open the Math Assistant pane. Learn more: Create your equation using ink or text. Step 2: Solve your equation To solve the current equation, do any of the following:

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