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The whole reason for solving an equation is to try to work something out. In questions, this “thing” that you are trying to work out is often represented by a **variable** such as **$x$ or $y$**. However, this is just shorthand for an unknown quantity. $x$ could represent the cost of apples in a supermarket, the age of Jack's sister, or even an unknown angle in a shape. In this article, we will not only be solving Equations but forming equations to show us how useful solving equations can actually be. The process of forming an equation is called **deriving** an **equation**.

## Deriving Equations Meaning

We solve Equations a lot but what actually is an equation? If we break down the word, we get equa+tion… ‘Equa’ looks a bit like equal. Thus, an equation is essentially anything with an **equal** **sign;** it is a statement of equality between two variables. So, if we are given a wordy question involving the equality of certain variables, we can form and solve an equation.

In mathematics, the process of forming a mathematical equation or formula is called **deriving**. We say we derive an equation to help us work something out. In the below section, we will be deriving equations and solving them to work out an unknown quantity.

A **variable** is some kind of **letter** or **symbol** standing for an **unknown** value. We often define $x$ and $y$ for variables however it can be any letter or symbol that represents an unknown quantity.

### Methods for Deriving an Equation

**1. Define Variables**

To derive an equation, first **define** any **unknown** **variables** to establish what you are actually trying to work out. For example, if the question asks you to work out the age of someone, define the person's age as a letter such as $x$. If the question asks you to work out the cost of something, define the cost to be some variable such as $c$.

**2. Identify Equal Quantities **

The next step is to work out where the **equals** **sign** goes. This might be explicitly stated in the question, for example, "the sum of the boy's ages is **equal** to 30." or "the cost of three apples **is** $30p$". However, sometimes it is less obvious and you have to use your imagination a little. For example, if we have three unknown Angles on a straight line, what do we know? The sum of angles on a straight line is **equal** to 180 degrees so we could use this. If we have a square or rectangle, we know that the parallel sides are **equal**, and so we could also use this. In the examples in the questions below, we will go through lots of common types of questions that involve deriving equations.

## Deriving Equations Examples

In this section, we will look at a range of different types of questions involving deriving equations. If you follow along, this should give you plenty of practice in deriving equations.

### Finding Missing lengths and Angles

**On the straight line below, work out the value of angle DBC. **

**Solution:**

Here we have a straight line with missing angles. Now, we know that the sum of angles on a straight line is equal to 180 degrees. Therefore, we can say $2a+3+90+6a-1=180$. By collecting like terms, we can simplify this to $8a+92=180$. Thus, we have just derived an equation! Now we can solve this equation to work out what a is, and plug this into the missing angles to identify the size of each of the angles.

Subtracting 92 from both sides, we get $8a=88$. Finally, dividing both sides by 8, we get $a=11.$

Thus, angle ABE=$2\times 11+3=25\xb0$, angle EBD we already know is 90 degrees, and angle DBC=$6\times 11-1=65\xb0$. Answering the original question, angle DBC is 65 degrees.

**Below is a rectangle. Work out the area and perimeter of this rectangle. **

**Solution:**

Since we have a rectangle, we know that the two parallel sides are the same. Thus, we could say that AB is equal to DC and thus $2x+15=7x+5$. We have therefore again derived another equation. To solve this equation, first subtract $2x$ from both sides to get $15=5x+5$. Then subtract five from both sides to get $10=5x$. Finally divide both sides by 5 to get $x=2$.

Now that we know the value of $x$, we can work out the lengths of each of the sides of the rectangle by substituting in $x$ into each of the sides. We get that the sizes of AB and DC are $2\times 2+15=19cm,$ and the lengths of AD and BC are $3\times 2=6cm.$ Since the perimeter is the sum of all of the measurements, the perimeter is $19+19+6+6=50cm.$Since the area is $base\times height$, we get that the area is $19\times 6=114c{m}^{2}$.

**The height of triangle ABC is $\left(4x\right)cm$, and the base is $\left(5x\right)cm$. The area is $200c{m}^{2}$. Work out the value of $x$.**

**Solution:**

Since the height is $4x$ and the base is $5x$, the area is $\frac{1}{2}\times 5x\times 4x=10{x}^{2}$. Now, we know that the area is $200c{m}^{2}$. Thus, $10{x}^{2}=200$ and so${x}^{2}=20$ and so $x=\sqrt{20}=4.47cm$

**Work out the size of the largest angle in the below triangle.**

**Solution:**

Since angles in a triangle sum to 180 degrees, we have $3x+5+6x+7+8x-2={180}^{\xb0}$. Simplifying, we could say $17x+10={180}^{\xb0}$. Therefore, we have derived another equation, and now we just need to solve it to work out x.

Subtracting ten from both sides, we get $17x={170}^{\xb0}.$Finally, dividing both sides by 17, we obtain $x={10}^{\xb0}$.

Since we have now found x, we can substitute it into each angle to find the largest angle.

Angle BAC= $6\times 10+7=67\xb0$

Angle ACB= $8\times 10-2=78\xb0$

Angle CBA= $3\times 10+5=35\xb0$

Thus, angle ACB is the largest and it is 78 degrees.

**Work out the size of angle ABD below. **

**Solution:**

Since opposite angles are **equal**, we know that $11x+2=13x-2$

To solve this, first subtract $11x$ from both sides to obtain $2=2x-2$. Then add 2 to both sides to get $4=2x$. Finally divide both sides by 2 to get $x=2$.

Substituting $x=2$ back into the angles, we have that angle ABD= $11\times 2+2=24\xb0$. Since angles on a straight-line sum to 180, we also get that angle ABC=$180-24=156\xb0$

**In the below diagram, the square has a perimeter twice that of the triangle. Work out the area of the square. **

**Solution:**

The perimeter of the triangle is $2x+3x+2x+3$ which can be simplified to $7x+3$. All the sides of the square are the same and so the perimeter is $5x+5x+5x+5x=20x.$ The perimeter of the square is twice that of the triangle, we have $2(7x+3)=20x$. If we expand the brackets, we get $14x+6=20x$. Subtracting $14x$ from both sides, we get $6=6x$ and divide both sides by six we finally obtain $x=1$. Thus, the length of the square is five units and the area of the square is $5\times 5=25uni{t}^{2}$

### Word Equations

**Catherine is 27 years old. Her friend Katie is three years older than her friend Sophie. Her friend Jake is twice as old as Sophie. The sum of their ages is 90. Work out Katie's age.**

**Solution:**

The first thing to acknowledge is that this question does not have many real-life applications, and it is more of a riddle than anything else. You could just ask each of Catherine's friends how old they are in real life, but that would be far less fun. It does provide us with some practice with forming and solving equations, so let us start by defining Sophie's age to be $x$.

If Sophie is $x$ years old, Katie must be $x+3$ years old since she is three years older than Sophie. Jake must be $2x$years old since he is twice Sophie's age. Now, since all of the sum of their ages to $90$, we have $27+x+x+3+2x=90$. Simplifying this, we get $4x+30=90$. Subtracting 30 from both sides, we get $4x=60$ and dividing both sides by four, we get $x=15$.

Thus, Sophie is 15 years old, so Katie must be $15+3=18$years old.

**The cost of a tablet is $\pounds x$. A computer costs $\pounds 200$ more than a tablet. The price of the tablet and computer is $\pounds 2000$. Work out the cost of the tablet and computer. **

**Solution:**

First, the tablet has already been defined to be $x$ pounds. The cost of the computer is $x+200$. Since the tablet and computer cost is $\pounds 2000$, we can say that $x+x+200=2000$. Simplifying, we get $2x+200=2000$. Thus we can solve this to find the price of the tablet.

Subtracting $200$ from both sides, we get $2x=1800$ and then dividing both sides by two$x=900.$ Thus, the tablet costs $\pounds 900$ and the computer cost$900+200=\pounds 1100$.

**Annabelle, Bella and Carman each play some games of dominoes. Annabelle won 2 more games than Carman. Bella won 2 more games than Annabelle. Altogether, they played 12 games, and there was a winner in every game. How many games did each of them win? **

Solution:

Again, we could just look at the score sheet in real life. However, for this exercise, we will form and solve an equation...

Define the Number of games Carman won to be $x$. Thus Annabelle won $x+2$ games, and Bella won $x+2+2$ games. So Bella won $x+4$ games. Altogether they played $12$ games, and there was a winner in every game, thus $x+x+2+x+4=12$. Simplifying this, we get $3x+6=12$. Subtracting six from both sides $3x=6$ and dividing both sides by 3, we get $x=2$. Therefore, Annabelle won 4 games, Bella won 6 games and Carman won 2 games.

## Deriving Equations - Key takeaways

- An equation is a
**statement**with an**equal****sign**. - In mathematics, forming a mathematical equation or formula is called
**deriving**. - We can derive equations when we know two quantities are equal.
- Once we have derived an equation, we can solve this equation to find an unknown variable.

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##### Frequently Asked Questions about Deriving Equations

What is the meaning of deriving equation?

It means to form an equation to help us to find some kind of unknown quantity.

What is an example of deriving an equation?

Suppose a multipack of beans in the supermarket costs £1 and beans come in a pack of four. If each of the tins of beans cost x pounds, we could derive an equation to say that 4x=1 and so by solving this, we get that x=0.25. In other words, each of the tins of beans cost 25p.

What are the methods for deriving an equation?

Define the variable you are trying to work out as a letter, for example, x. Then work out where equality holds and put an equals sign in the equation where necessary.

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