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Resources to Help You Solve Math Equations

Whether you love math or suffer through every single problem, there are plenty of resources to help you solve math equations. Skip the tutor and log on to load these awesome websites for a fantastic free equation solver or simply to find answers for solving equations on the Internet.

Stand By for Automatic Math Solutions at Quick Math

The Quick Math website offers easy answers for solving equations along with a simple format that makes math a breeze. Load the website to browse tutorials, set up a polynomial equation solver, or to factor or expand fractions. From algebra to calculus and graphs, Quick Math provides not just the answers to your tough math problems but a step-by-step problem-solving calculator. Use the input bar to enter your equation, and click on the “simplify” button to explore the problem and its solution. Choose some sample problems to practice your math skills, or move to another tab for a variety of math input options. Quick Math makes it easy to learn how to solve this equation even when you’re completely confused.

Modern Math Answers Come From Mathway

Mathway offers a free equation solver that sifts through your toughest math problems — and makes math easy. Simply enter your math problem into the Mathway calculator, and choose what you’d like the math management program to do with the problem. Pick from math solutions that include graphing, simplifying, finding a slope or solving for a y-intercept by scrolling through the Mathway drop-down menu. Use the answers for solving equations to explore different types of solutions, or set the calculator to offer the best solution for your particular math puzzle. Mathway offers the option to create an account, to sign in or sign up for additional features or to simply stick with the free equation solver.

Wyzant —​ Why Not?

Wyzant offers a variety of answers when it comes to “how to solve this equation” questions. Sign up to find a tutor trained to offer online sessions that increase your math understanding, or jump in with the calculator that helps you simplify math equations. A quick-start guide makes it easy to understand exactly how to use the Wyzant math solutions pages, while additional resources provide algebra worksheets, a polynomial equation solver, math-related blogs to promote better math skills and lesson recording. Truly filled with math solutions, Wyzant provides more than just an equation calculator and actually connects you with people who are trained to teach the math you need. Prices for tutoring vary greatly, but access to the website and its worksheets is free.

Take in Some WebMath

Log onto the WebMath website, and browse through the tabs that include Math for Everyone, Trig and Calculus, General Math and even K-8 Math. A simple drop-down box helps you to determine what type of math help you need, and then you easily add your problem to the free equation solver. WebMath provides plenty of options for homeschoolers with lesson plans, virtual labs and family activities.

Khan Academy Offers More Than Answers

A free equation solver is just the beginning when it comes to awesome math resources at Khan Academy. Free to use and filled with videos that offer an online teaching experience, Khan Academy helps you to simplify math equations, shows you how to solve equations and provides full math lessons from Kindergarten to SAT test preparation. Watch the video for each math problem, explore the sample problems, and increase your math skills right at home with Khan Academy’s easy-to-follow video learning experience. Once you’ve completed your math video, move onto practice problems that help to increase your confidence in your math skills.

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Problem Solver Subjects

Our math problem solver that lets you input a wide variety of math math problems and it will provide a step by step answer. This math solver excels at math word problems as well as a wide range of math subjects.

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Here are example math problems within each subject that can be input into the calculator and solved. This list is constanstly growing as functionality is added to the calculator.

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Below are examples of basic math problems that can be solved.

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Math Word Problem Solutions

Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms.

Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 × b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

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Precalculus : Application Problems

Study concepts, example questions & explanations for precalculus, all precalculus resources, example questions, example question #1 : application problems.

The equation of the value for this problem is 

We can divide by R to get

We want to solve for n in this case, which is the amount of years. If we use the natural log on both sides and properties of logarithms, we get

If we solve for n, we get

Since we are investing for two years with a yearly rate of 5%, we will use the formula to calculate compound interest.

Our amount after two years is:

Example Question #3 : Application Problems

Plugging this into the equation for compound interest gives us the expression:

Example Question #4 : Application Problems

Plugging our numbers into the formula for compound interest, we have:

After placing his money into the other savings account, he has

Example Question #5 : Application Problems

None of the other answers.

The formula for the compund interest is as follows:

By substuting known values into the compound interest formula, we have:

From here, substitute known values.

Example Question #6 : Application Problems

The formula is as follows:

Substitute known values.

Take the natural log of both sides.

Use the log power rule.

Example Question #7 : Application Problems

The amount of phosphorus present in a sample at a given time is given by the following equation:

The problem asks us for the percent that the amount of phosphorus after t days is of the original amount of phosphorus. If we think of this percentage with respect to the variables present in the equation, we can see that the following fraction expresses the amount of phosphorus after t days as a percentage of the initial amount of phosphorus:

So after 25 days of decay, the amount of phosphorus is 47% of the initial amount.

Example Question #8 : Application Problems

The exponential decay of an element is given by the following function:

Example Question #9 : Application Problems

The exponential decay of an element is given by the function:

Substituting in the values from the problem gives

Example Question #10 : Application Problems

The exponential decay of an element is given by the function

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• ISBN-10 0878915567
• ISBN-13 978-0878915569
• Edition Revised
• Publisher Research & Education Association
• Publication date October 26, 1984
• Language English
• Dimensions 6.75 x 2 x 9.75 inches
• Print length 1104 pages
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From the back cover, about the author, excerpt. © reprinted by permission. all rights reserved., product details.

• Publisher ‏ : ‎ Research & Education Association; Revised edition (October 26, 1984)
• Language ‏ : ‎ English
• Paperback ‏ : ‎ 1104 pages
• ISBN-10 ‏ : ‎ 0878915567
• ISBN-13 ‏ : ‎ 978-0878915569
• Reading age ‏ : ‎ 16 years and up
• Item Weight ‏ : ‎ 3.09 pounds
• Dimensions ‏ : ‎ 6.75 x 2 x 9.75 inches
• #38 in Popular & Elementary Pre-Calculus (Books)
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About the author

Dennis c. smolarski.

I am a Professor of Mathematics and Computer Science at Santa Clara University (California). My undergraduate studies were at Santa Clara University (1965-69). My graduate work was completed at the University of California at Santa Barbara (1973-75) and at the University of Illinois (1979-1982).

As a Catholic priest and a member of the Jesuit Order (with theological studies completed at the Jesuit School of Theology at Berkeley, 1976-1979), I also have an interest in religion, particularly in liturgy and worship, and have numerous publications in that field as well.

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Chapter 6.8: Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized,  use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips,  put this information in the distance column. Now use this table to set up and solve the following examples.

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

• A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
• Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
• Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
• Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
• A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
• Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
• A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
• A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

• A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
• A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
• A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
• As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
• Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
• A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
• A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
• A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
• Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
• Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
• Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
• Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
• On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
• Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answers to odd questions

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Precalculus word problem

An arch going into an alley is in the shape of a downward opening parabola. The base of the arch is 10 ft. wide, and the arch is 15 ft. high at its center. How wide is the arch 10 ft. up?

2 Answers By Expert Tutors

Joe H. answered • 12/09/19

Certified Advanced Math Teacher

Since the arch is a downward facing Parabola and you're given the width of the base (10ft) as well as the height at the center (15ft) you can use the Vertex Form equation for a parabola y=a(x-h)^2+k .

To help, you can draw an xy-axis then label your x-intercepts at the points (0,0) and (10,0), then your vertex is at the point (5,15).

To find the correct equation you'll need to solve for a. So, plug in the coordinates of your vertex (5,15) as h=5 and k=15. Now pick either one of your intercept points, for example I used the point (0,0) and substitute the values into your Vertex Form equation like so 0=a(0-5)^2+15. Now, simplify and solve for a and you should get a=-3/5.

Next, to find the width at 10ft up you'll need to first set the equation of your parabola equal to 10 then solve for x 1 and x 2 .

10 = -3/5 * (x-5)^2 +15 which yields the values x 1 =2.113 and x 2 =7.887.

Finally, the width of the arch relative to those x-values is found by simply adding the distance from the center (x=5) to each x 1 =2.113 and x 2 =7.887. Do this once, then use symmetry

5 - 2.113 = 2.887 therefore the distance between x 1 and x 2 is roughly 5.774 ft.

Therefore the width of the arch 10 feet up is about 5.8 ft.

Mark M. answered • 12/09/19

Mathematics Teacher - NCLB Highly Qualified

Draw and label a diagram!

Place vertex at (0, 15), x-intercepts at (-5, 0) and (5, 0)

f(x) = a(x - 0) 2 + 15

f(x) = ax 2 + 15

0 = a(5) 2 + 15

f(x) = -3/5(x - 0) 2 + 15

10 = -3x 2 /5 + 15

-5 = -3x 2 /5

-25 = -3x 2

±(5√3)/3 = x

Distance between arches at 10 feet up is 10√3 / 3 feet.

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