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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.


problem solving solving equations

Chapter 1: Solving Equations and Inequalities

Problem solving, learning objectives.

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

The power of algebra is how it can help you model real situations in order to answer questions about them.

Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started.

Twenty-eight less than five times a certain number is 232. What is the number?

Following the steps provided:



In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.

For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x , is [latex]x+1[/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.

We apply the idea of consecutive integers to solving a word problem in the following example.

The sum of three consecutive integers is 93. What are the integers?


Combine like terms, simplify, and solve.

[latex]\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}[/latex]

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:


Likewise, if we want to find rate, we can isolate r using division:


In the following examples you will see how this formula is applied to answer questions about ultra marathon running.

Ann Trason

Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993 Trason finished the run with a time of 6:09:08.  The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21. [2]

In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].

(rounded to two decimal places)

Johnson’s rate:

[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , and the time Trason had run is 5.5 hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours.  What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.

Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.

Now we will tackle the last question where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t , the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P ), which then grows slowly according to the interest rate ( R , measured in percent) and the length of time ( T , usually measured in months) that the money stays in the account. The amount earned over time is called the interest ( I ), which is then given to the customer.


The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of $2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite 0.7% as the decimal 0.007, then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has $2336 after 24 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]

Define and Translate:  we know P, R, and T so we can use substitution. R  = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years.  We need to change T into months because we can’t change the rate—it is set by the bank.

[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned $756.

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested $1000.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand:  We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

First we will substitute the dimensions we know into the formula for perimeter:


Then we will isolate w to find the unknown width.


Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

If we replace the width we found, [latex]w=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:


Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  


First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.


The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where

A = area b = the length of the base h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Find the base ( b) of a triangle with an area ( A ) of 20 square feet and a height ( h) of 8 feet.

Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex] .

Substitute the given lengths into the formula and solve for  b.

[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]

The base of the triangle measures 5 feet.

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x .

Use the multiplication and division properties of equality to isolate the variable b .


Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h .



Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].


Add 32 to both sides.



Think About It

Express the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex], in terms of the height, h .

In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h .

[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].

[latex]\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

Check Your Solutions

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 − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , 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!

Download on App Store

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The equations section of QuickMath allows you to solve and plot virtually any equation or system of equations. In most cases, you can find exact solutions to your equations. Even when this is not possible, QuickMath may be able to give you approximate solutions to almost any level of accuracy you require. It also contains a number of special commands for dealing with quadratic equations.

The Solve command can be uses to solve either a single equation for a single unknown from the basic solve page or to simultaneously solve a system of many equations in many unknowns from the advanced solve page . The advanced command allows you to specify whether you want approximate numerical answers as well as exact ones, and how many digits of accuracy (up to 16) you require. It also allows you to eliminate certain variables from the equations.

Go to the Solve page

The Plot command, from the Graphs section, will plot any function of two variables. In order to plot a single function of x, go to the basic equation plotting page , where you can enter the equation and specify the upper and lower limits on x that you want the graph to be plotted for. The advanced plotting page allows you to plot up to 6 equations on the one graph, each with their own color. It also gives you control over such things as whether or not to show the axes, where the axes should be located, what the aspect ratio of the plot should be and what the range of the dependent variable should be. All equations can be given in the explicit y = f(x) form or the implicit g(x,y) = c form.

Go to the Equation Plotting page

The Quadratics page contains 13 separate commands for dealing with the most common questions concerning quadratics. It allows you to : factor a quadratic function (by two different methods); solve a quadratic equation by factoring the quadratic, using the quadratic formula or by completing the square; rewrite a quadratic function in a different form by completing the square; calculate the concavity, x-intercepts, y-intercept, axis of symmetry and vertex of a parabola; plot a parabola; calculate the discriminant of a quadratic equation and use the discriminant to find the number of roots of a quadratic equation. Each command generates a complete and detailed custom-made explanation of all the steps needed to solve the problem.

Go to the Quadratics page

Introduction to Equations

By an equation we mean a mathematical sentence that states that two algebraic expressions are equal. For example, a (b + c) =ab + ac, ab = ba, and x 2 -1 = (x-1)(x+1) are all equations that we have been using. We recall that we defined a variable as a letter that may be replaced by numbers out of a given set, during a given discussion. This specified set of numbers is sometimes called the replacement set. In this chapter we will deal with equations involving variables where the replacement set, unless otherwise specified, is the set of all real numbers for which all the expressions in the equation are defined.

If an equation is true after the variable has been replaced by a specific number, then the number is called a solution of the equation and is said to satisfy it. Obviously, every solution is a member of the replacement set. The real number 3 is a solution of the equation 2x-1 = x+2, since 2*3-1=3+2. while 1 is a solution of the equation (x-1)(x+2) = 0. The set of all solutions of an equation is called the solution set of the equation.

In the first equation above {3} is the solution set, while in the second example {-2,1} is the solution set. We can verify by substitution that each of these numbers is a solution of its respective equation, and we will see later that these are the only solutions.

A conditional equation is an equation that is satisfied by some numbers from its replacement set and not satisfied by others. An identity is an equation that is satisfied by all numbers from its replacement set.

Example 1 Consider the equation 2x-1 = x+2

The replacement set here is the set of all real numbers. The equation is conditional since, for example, 1 is a member of the replacement set but not of the solution set.

Example 2 Consider the equation (x-1)(x+1) =x 2 -1 The replacement set is the set of all real numbers. From our laws of real numbers if a is any real number, then (a-1)(a+1) = a 2 -1 Therefore, every member of the replacement set is also a member of the solution set. Consequently this equation is an identity.  

problem solving solving equations

The replacement set for this equation is the set of real numbers except 0, since 1/x and (1- x)/x are not defined for x = 0. If a is any real number in the replacement set, then

problem solving solving equations

Math Topics

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Unit 2: Lesson 1

Equations with variables on both sides

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

7th Grade Math Problems   8th Grade Math Practice   From Word Problems on Linear Equations to HOME PAGE

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