problem solving maths equations

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.

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.

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:

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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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

Every week, we teach lessons on solving equations to students in schools and districts across the US as part of our online one-on-one math tutoring programs. On this page we’ve broken down everything we’ve learnt about teaching this topic effectively.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

[FREE] Solving Equations Worksheet (Grade 8)

Use this worksheet to check your 8th grade students’ understanding of solving equations. 15 questions with answers to identify areas of strength and support!

Common Core State Standards

How does this relate to 8 th grade and high school math?

• Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

• Combine like terms .
• Simplify the equation by using the opposite operation to both sides.
• Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

• Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
• Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
• Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
• Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

• Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
• Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
• Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
• Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
• Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solve equations with fractions
• Linear equations
• Writing linear equations
• Substitution
• Identity math
• One step equation

Practice solving equations questions

1. Solve 4x-2=14.

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

• Inequalities
• Types of graph
• Math formulas
• Coordinate plane

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Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

• Read the whole thing first
• Do a sketch if possible
• Assign letters for the values
• Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

• Let S = dollars Sam has
• Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

• Let D = number of dogs
• Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

• Use w for width of rectangle: w = 12m
• Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

• Use S for how many games Sam played
• Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

• Use a for Alex's work rate
• Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

• The number of "5 hour" days: d
• The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

Example: A circle has an area of 12 mm 2 , what is its radius?

• Use A for Area: A = 12 mm 2
• Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

• Use V for Volume
• Use A for Area
• Use s for side length of cube
• Volume of a cube: V = s 3
• Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

• Joel's normal rate of pay: $N per hour
• Joel works for 40 hours at $N per hour = $40N
• When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
• Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
• And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

• Present Value PV = $2,000
• Interest Rate (as a decimal): r = 0.11
• Number of Periods: n = 3
• Future Value (the value we want): FV

We are being asked for the Future Value: FV

Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

• Present Value PV = $1,000
• Interest Rate (the value we want): r
• Number of Periods: n = 9
• Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

• Number of boys now: b
• Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

• We could try, say, n=10: 10(12) = 120 NO (too small)
• Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

• the length of the room: L
• the width of the room: W
• the total Area including veranda: A
• the width of the room is half its length: W = ½L
• the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

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Brain Teaser: Think You're a High IQ Individual? Solve This Math Puzzle to Prove It!

Brain Teaser: It’s a tough challenge! Only the brightest high-IQ individuals can beat this math puzzle. Are you one of them? You have 19 seconds on the clock to beat this challenge!

Brainteasers are more than just riddles wrapped in a package. They're a mental gym, a challenging playground where you can push your cognitive abilities. 

These puzzles come in all shapes and sizes, like a well-stocked toy box for the mind.

From logic puzzles that test your reasoning skills to wordplay challenges that make you think outside the box, there's a brainteaser out there for everyone, regardless of age or background.

Try: Wordmoji Puzzle: Can Your Emoji Mastery Crack This Chocolate Mystery in 9 Seconds?

But what exactly makes a brainteaser a brainteaser?  It all comes down to the unconventional thinking they require. 

This particular brainteaser is a siren song for all math enthusiasts! Get ready to flex your cognitive muscles and test your mathematical prowess.  

Below, you'll find an image featuring a monkey and various other objects with numerical values assigned to them. Your mission is to decipher the value written against the objects and decode the final equation mentioned in the image. 

Here's the catch:  you only have 19 seconds!  The pressure is on, but fear not, math whizzes!  

By utilising lateral thinking and approaching the problem from a creative angle, you'll be well on your way to cracking the code.

Are you ready to embark on this mathematical adventure?  Set your timer, focus your mind, and let the brainteasing begin!  

Remember, the key lies in unconventional thinking. 

Brain Teaser: Find the Values in 19 Seconds

Source: Pinterest (Vera Pavlovskaya) 

Brainteasers are more than just mind games; they're a playground for your brain, offering a delightful blend of challenge and amusement.  

Not only can they keep your mind sharp, but they can also provide an opportunity to impress your friends with your problem-solving prowess.

The brainteaser we're about to tackle involves deciphering various equations.  This isn't your ordinary math problem.  

It's a test of both logic and creative thinking.  

By examining the equations provided and utilising your basic arithmetic skills, you'll be well on your way to unlocking the secret.

Time is of the essence!  Being under pressure can actually benefit your brain.  

It forces you to focus your attention and think more quickly, honing your ability to recognise patterns and make logical connections.  

But fear not, even if the clock runs out before you crack the code, all is not lost!

So, how many equations did you solve?  Congratulations if you managed to find them all!  Your logical skills deserve a round of applause.  

If you're still stumped, don't worry!  The beauty of brainteasers is that you can always revisit them and try again.  Simply scroll back up and give it another shot.

Find the Values: Solution 

The image shows two Monkeys valued at 6 in the first equation

Let's take the value of one Monkey to be A

So, we get: 

A=6/2 

So the value of one Monkey is 3 

Coming to the second equation we have a Monkey and two bananas subtracting and giving the value of 13

Let's take the value of Bananas to be B

So, the value of Bananas is 16 

Coming to the third equation we have Bananas and a Coconut Tree valued at 20 

Let's take the value of Coconut Tree to be C

So we have: 

B+C=20 

16+C=20 

Coming to the fourth equation we have 

Deriving the values from the above we get: 

3+4=7 

So, the value of the final equation is 7

Hope you liked this puzzle! Keep trying your hands on these puzzles and you will surely emerge as a true puzzle master! 

Must Try: Logic Puzzle: Only High IQ Thinkers Can Find the Hidden Word in 13 Seconds—Will You?

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• \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
• \mathrm{Kira\:went\:for\:a\:drive\:in\:her\:new\:car.\:She\:drove\:for\:142.5\:miles\:at\:a\:speed\:of\:57\:mph.\:For\:how\:many\:hours\:did\:she\:drive?}
• \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
• \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
• \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
• How do you solve word problems?
• To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
• How do you identify word problems in math?
• Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
• Is there a calculator that can solve word problems?
• Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
• What is an age problem?
• An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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• High School Math Solutions – Inequalities Calculator, Exponential Inequalities Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential...

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