Imagine that you want to solve the following problem:

Three friends, each with a bag of coins, exchange coins as follows: Ulf gives Stuart and Erik enough coins to double their holdings. Stuart then gives Ulf and Erik enough coins to double their holdings. Finally, Erik gives Ulf and Stuart enough coins to double their holdings. Erik had 36 gold coins from the beginning and 36 gold coins after the last transaction. How many gold coins do the three friends have in total? (Pythagoras Quest, District Final 2012, Part 1 Problem 6)

How would you do that? The problem contains various unknown values: we do neither know how many coins Ulf has, nor how many gold coins Stuart has. It can seem impossible to solve the problem since so many values are unknown, but many mathematical problems are similar to this one. To find unknown values is something recurring in mathematics, and there are methods that can help you solve such problems. In this lesson, you will learn about these methods. We will start by looking at how we can handle unknown numbers with the help of something that is called a variable. 

When solving mathematical problems, we often have to do calculations with an unknown number. For example, an exercise can lead to a calculation that looks like this: $[] - 2 = 4 $. The empty box means that we have an unknown number, and our goal is to find this number. When we do that, we can reason as follows: the unknown number must be $2$ greater than $4$, since if we subtract $2$ from it, we get $4$. Therefore, the unknown number must be $4 + 2 = 6$. If we replace the empty box with $6$, we get $6 - 2 = 4 $. That is true, and then we know that the empty box denotes the number $6$. 

As we can see in Exercise 1, we have to write a lot to explain how we find the unknown value. We have to write “the unknown number” or “the empty box” all the time so that the person looking at our solution understands what we mean. Furthermore, it can get very messy if an exercise contains various unknown numbers. If we denote all of them by an empty box, we won’t be able to keep track of which one is which. Therefore, we use something that is called variables.

A variable is a number that has an unknown value. Variables are commonly denoted by letters. Usually, we use x and y, but they could be named a, b, c, etc. as well. The advantages of using variables are that it takes less time to write a letter than it takes to draw an empty box, and that we can use different variables to denote different unknown numbers. 

Now that we know that variables are numbers with unknown values, we have to find a word for numbers with known values. They are called constants or constant numbers. All the numbers on the number line are constants. A bit plainly, we can say that a constant is the “opposite” of a variable, since a constant has a definite value, whereas the value of a variable is indefinite. 

Let’s look at exercise 1a) again. There, we wanted to find the unknown number that the empty box in $45 + [] = 61$ denotes. Now that we know that variables are used to denote unknown numbers, we can replace the empty box with “$x$”. Then we write $45 + x = 61$, and our goal is to find the value of $x$. We can do that as follows: 

$x$ must be $45$ less than $61$, since we have to add $45$ to $x$ to get $61$. Therefore, $x$ must be $61 - 45 = 16$. 

Answer: $\mathbf{x = 16}$.

If we do this, we have to write much less, and the calculations become much clearer, since we can always write $x$ instead of “the unknown number” or “the empty box”.

Let’s look at the problem with the three friends again. We know that Erik had $36$ coins in the beginning, but the amount of coins that Ulf and Stuart had from the beginning is unknown. Then, we can denote these amounts by variables: $u$ can denote the amount of coins that Ulf had from the beginning, and $s$ can denote the amount that Stuart had from the beginning. We know that Ulf gives Stuart and Erik enough coins to double their holdings. But how can we write that? How can we do calculations with variables?

Variables are numbers, or more specifically, they denote numbers. What does this say about the calculation rules for variables? Do you think that the same rules could apply to both variables and constant numbers? If you answer yes to that question, you are absolutely right, and in this section, you will learn more about how to use these rules. 

When we add the same number to itself multiple times, we do a multiplication. For example, $3 + 3 + 3 + 3$ can be written as $3 \cdot 4$. We can do the same thing with variables: for example, $x + x + x + x$ is the same as $x \cdot 4$. 

Multiplication of numbers is commutative. That means that we can multiply the same numbers in whichever order we want, and still get the same product. For example, $3 \cdot 2 = 2 \cdot 3 = 6$. The same goes for variables. For instance, $x \cdot 3 = 3 \cdot x$, where $x$ denotes any number. 

Something that is different for variables is that we can omit the multiplication sign between a constant and a variable. So, we can for example write $3a $ instead of $3 \cdot a$ or $4x$ instead of $x + x + x + x$.

$4x$ equals $x4$, since multiplication of numbers is commutative, but we usually write such products as $4x$ and not as $x4$. 

Variables can be added in the same way as constant numbers. For example, $14 + 16 = 30 $. In the same way, $14x + 16x = 30x $. We don’t know the value of $x$, but if we have got $14$ pieces of $x$ and $16$ pieces of $x$, that makes $30$ pieces of $x$ in total. 

Subtraction of variables works the same way. Just like $30 - 16 = 14$, $30x - 16x = 14x$.

Earlier, we have looked at how an addition of variables can be written as a multiplication. We can also examine what happens when variables and constant numbers are multiplied by each other. We already know that $3 \cdot x = 3x $. But what happens when we for example want to calculate $5 \cdot 3x $? Than, we can do as follows:

$ 5 \cdot 3x = 5 \cdot 3 \cdot x = 15 \cdot x = 15x $

When we do such calculations, we have to remember that multiplication of both constants and variables is commutative. It doesn’t matter how many factors we have and if they are constants or variables. As long as we are only doing multiplication, we can write the factors in any order we want without changing the value of the product. So, for example we have:

$ 5 \cdot x \cdot 3 = 5 \cdot 3 \cdot x = 15 \cdot x = 15x $

When we do calculations with constants, we can’t divide by $0 $. So what happens if we divide a constant or a variable by another variable? Take $ \frac {10}{x} $ for example. Since $x $ has an unknown value, we don’t know whether or not $x = 0 $, and it is possible that we divide $10 $ by $0 $ if we divide it by $x $. So, if we don’t want our calculations to be wrong, we shouldn’t divide by $x $ if we aren’t certain that the value of $x $ isn’t $0 $. 

Now that you know a bit about calculations with one variable, we can go further with the exercise from _1. We know that Ulf gives Stuart and Erik enough coins to double their holdings. Then, Ulf will give away as many coins as Stuart and Erik have in total. He will then have $u - s - 36 $ coins left. At the same time, Stuart will have $2s $ coins and Erik will have $36 \cdot 2 = 72 $ coins after they have doubled their holdings. 

After that, Stuart gives Ulf and Erik enough coins to double their holdings. Then, Stuart will give away as many coins as Ulf and Erik have in total, and he will have $2s - (u - s - 36) - 72 $ coins left. Now we have a calculation with several variables. How is that supposed to be done?

There are mathematical problems with several unknown values, where we use different variables for different unknown numbers. In those cases, we have to do calculations with different variables. How do you think that is supposed to be done? Can we for example write $2s - (u - s -36) - 72 $ in a simpler way? We can, and in this section, you will learn how!

That addition and multiplication are commutative is a calculation rule for numbers. Since variables denote numbers, this rule applies to them too. For example, $x + y = y + x $ and $x \cdot y = y \cdot x$, where $x $ and $y $ denote any number. Because of that, calculation rules are often written using variables so that we understand that they apply to any number. That “multiplication of numbers is commutative” can be written as $a \cdot b = b \cdot a $, where the variables $a $ and $b $ denote arbitrary numbers. Since the multiplication sign is commonly omitted when multiplying with variables, we can write $ab = ba $. 

Even though $ab = ba $, we commonly write the variables in alphabetical order. That means that we usually write $ab $ and not $ba $, even though both are correct and have the same value. 

When we work with expressions with variables, we want to write them in as simple and as short a way as possible. That is, we want to simplify the expressions. The simplification must apply to all possible values of the variables for it to be valid. An example of an expression that can be simplified is $2x + 3x $. It can be written as $5x $. $2x + 3x = 5x $ is always true, no matter the value of $x $. 

Our goal is usually to simplify expressions as far as possible. That means that we simplify them until they can’t be written in a simpler way. If we write $5x + 4x $ instead of $2x + 3x + 4x $, we haven’t simplified the expression as far as possible, since it can be simplified further according to $5x + 4x = 9x $.

$14x + 16y $ can, unlike $2x + 3x = 5x $ not be simplified. We can’t say anything about how much $14x + 16y $ is in general without knowing how the values of the variables relate to each other. They could have the same value, but they could just as well have different values, and then we can’t add them up. It’s like saying that we have $14 $ pencils and $16 $ candy bars. It is true that we have $30 $ objects in total, but pencils and candy bars are different things, and we can’t add them up to get $30 $ pencils or $30 $ candy bars. 

We can now go back to the exercise from 1. We have seen that Stuart has $2s - (u - s - 36) - 72 $ coins left after he has given Ulf and Erik enough coins to double their holdings. Now, we can simplify this and we can see that the amount of coins that Stuart has left is $2s - (u - s - 36) - 72 = 2s - u + s + 36 - 72 = 3s - u - 36 $. At the same time, Ulf will have $2(u - s - 36) = 2u - 2s - 72 $ coins and Erik will have $2 \cdot 72 = 144 $ coins after they have doubled their holdings. 

After that, Erik gives Ulf and Stuart enough coins to double their holdings. Then, Erik gives away as many coins as Ulf and Stuart have in total, and he will have $$ 144 - (2u - 2s - 72) - (3s - u - 36) = \\ = 144 - 2u + 2s + 72 - 3s + u + 36 = 252 - u - s $$ left. 

We know that Erik now has $36 $ coins, and thus $252 - u - s $ should equal $36 $. But what can we do with that knowledge, and how are we supposed to determine how many coins the friends have in total?

In Exercise 1, we wanted to find the value of $x $ with the help of the condition $45 + x = 61 $. Such an equality is called an equation. When it comes to equations, the values on both sides of the $= $-sign are equal. These values are called expressions. Expressions can be made up of constants, variables, or both. So, $45 + x $ and $61 $ are expressions, and when we connect them with an $= $-sign, we get an equation. 

The purpose of many mathematical problems is to find unknown values. By doing calculations with equations, one can solve such problems in a fast and simple way. Thus, equations are highly useful, and therefore, you will learn about them in this section. First, we will see how a problem can be solved with the help of an equation, and then we will dig deeper into what an equation is and see some more examples of calculations with equations. 

Olof thought of a number. He multiplied the number by $3 $ and subtracted $5 $ from the product. Then, he divided the result by $2 $, and got the same number as the one he thought of from the beginning. Which number did he think of?

Without any comments, the solution of the equation would have looked like this:

$$ \frac {3x - 5} {2} = x $$

$$ \frac {3x - 5} {2} \cdot 2 = x \cdot 2 $$

$$ 3x - 5 = 2x $$

$$ 3x - 5 + 5 = 2x + 5 $$

$$ 3x = 2x + 5 $$

$$ 3x - 2x = 2x - 2x + 5 $$

$$ x = 5 $$

We commonly begin a new line for each new calculation that we do. That’s practical, because it becomes easy to discover errors when comparing two lines with each other. 

In the example above, $\frac {3x - 5} {2} $ equals $x $. In that way, equations can be said to be statements, since this equation, for example, corresponds to the statement “$\frac {3x - 5} {2} $ is equal to $x $”. Equations commonly contain variables, but they don’t necessarily have to. $1 + 1 = 2 $ is also an equation, because it is a statement about the values on both sides of the $= $-sign being equal. Inequalities such as $4 < x $ are not equations, because they don’t concern values that are equal. 

The expression that stands on the right side of the $= $-sign in an equation is called the right-hand side, and the one that stands on the left side is called the left-hand side. We commonly abbreviate these to RHS and LHS. If a constant number replaces the variable in an equation so that LHS equals RHS, the constant number is a root of the equation. If we replace $x $ with $5 $ in the equation $\frac {3x - 5} {2} = x $, we get:

$$ LHS = \frac {3x - 5} {2} = \frac {3 \cdot x - 5} {2} = \frac {3 \cdot 5 - 5} {2} = 5 = RHS $$

Since $LHS = RHS $, $x = 5 $ is a root of the equation.

If a variable stands alone, it can have any value, but an equation can limit the number of values that a variable can have. If $x $ in the equation $x - 4 = 12 $ equals $1 $, we get $LHS = 1 - 4 = -3 $ and $RHS = 12 $. Thus, LHS and RHS are not equal, and $x $ can’t have the value $1 $. $x $ can actually not have any value except for $16 $ without LHS not being equal to RHS. That means that $x = 16 $ is the root of the equation, and that the equation limits the number of values that $x $ can have so that $x $ can’t have any other value than $16 $. 

With the help of the condition $LHS = RHS $, we can determine if a specific value of the variable in an equation is the root of the equation. We can use that if we want to check whether or not we have solved an equation correctly. 

Let’s try replacing $x $ with a different value than $5 $ in the equation $\frac {3x - 5} {2} = x$. We can have $x = 1 $ for instance. Instead of $\frac {3x - 5} {2} $ in LHS, we write $\frac {3 \cdot 1 - 5} {2} $, and we examine both sides:

$$ LHS = \frac {3 \cdot 1 - 5} {2} = -1 $$

$$ RHS = 5 $$

We can see that $-1 \neq 5 $, that is $LHS \neq RHS $. (The sign $\neq $ means “is not equal to”). The statement “$\frac {3 \cdot 1 - 5} {2} $ equals $1 $” is false, and $x = 1 $ is not a root of the equation. 

When we replace variables with constant numbers, it’s important to remember that for example $5x = 5 \cdot x $. Before replacing a constant number with a variable, it is wise to rewrite such products so that we have a multiplication sign between the terms. So, for example, $8y $ should be written as $8 \cdot y $. Otherwise, it is easy to make a mistake. Let’s, as an example, replace $y $ in $8y $ with $2 $: 

$$ 8y = 8 \cdot y = 8 \cdot 2 = 16 $$

If we don’t write $8y $ as $8 \cdot y $, there is a risk of doing this:

$$ 8y = 82 $$

That’s wrong, because $8y $ means that we have $8 $ pieces of $y $, $8 \cdot y $ that is, and not $8 $ times $10$ and one $y $ ($80 + y $). Earlier, we have seen that $8y = 16 $, which is another reason why $8y = 82 $ is wrong. 

Solve the equation $\frac {y} {3} + 1 = y - 5 $.

Solve the equation $100x + 10 = 200x - 90 $. 

Solve the equation $\frac {3a} {2} - 7 = 8 + \frac {a} {3} $.

In some cases, an equation has no roots, and in other cases, it can have infinitely many roots. You can identify an equation that lacks roots as follows: When solving the equation, you end up with a simple equation, for example $2 = 1 $, which is a false statement.

Solve the equation $x - 4 = x + 4 $.

If an equation has infinitely many roots, the variable can have any value, and the statement that the equation describes will still be true. You can see that by starting to solve the equation and then getting an equation that is valid for all values of the variable. Examples of such equations are $x = x $, $x + 4 = x + 4 $ and $0 = 0 $. In the first equations, we can replace $x $ with any value and still get $LHS = RHS$. No matter which value $x $ has, $x $ always equals $x $ and $x + 4 $ always equals $x + 4 $. The last equation doesn’t contain any variables, but if we simplify the first two ones, we get a similar equation. Let’s start with $x = x $. We subtract $x $ from both sides and get: 

$$ x = x $$

$$ x - x = x - x $$

$$ 0 = 0 $$

The same thing happens when we try to solve $x + 4 = x + 4 $ by subtracting $x $ and $4 $ from both sides:

$$ x + 4 = x + 4 $$

$$ x + 4 - 4 = x + 4 - 4 $$

$$ x = x $$

$$ x - x = x - x $$

$$ 0 = 0 $$

In both cases, we went from an equation that contains $x $ to an equation that doesn’t contain any variables ($0 = 0 $). We know that $0 $ always equals $0 $, and this tells us that the equation has infinitely many solutions. We can reason that whichever value $x $ has, $0 $ still equals $0 $. 

Equations are statements. A statement is something that is such that you can determine if it is true or false. If we have an equation that doesn’t contain variables, like $1 + 1 = 2 $ for example, you can determine if it’s true or false right away. In this case, the equation represents the statement “$1 + 1 $ equals $2 $”, which is true. 

If an equation contains one or more variable(s), like $x + 5 = 10 $ for example, you can’t determine if the statement that the equation represents is true or false right away. If the statement is true or false depends on the value of the variable, which can have any value. When we solve an equation, we look for the value that makes the statement true. That value of the variable is the root of the equation. 

When solving an equation, you write equivalent statements. That means that the statements are such that if one statement is true, the other one is also true, and the other way around. Let’s solve the equation $x + 5 = 10 $: 

$$ x + 5 = 10 $$

$$ x + 5 - 5 = 10 - 5 $$

$$ x = 5 $$

We get $x = 5 $. That $x + 5 = 10 $ and $x = 5 $ are equivalent statements means that if $x + 5 $ equals $10 $, than $x $ equals $5 $, and if $x $ equals $5 $, than $x + 5 $ equals $10 $. 

Let’s go back to the exercise from 1. Based on what we know, we can write the equation $252 - u - s = 36 $. But what should we do with an equation that contains more than one variable?

Sometimes, we encounter equations with multiple variables. So, how does that work? Can we find one solution to such an equation, or is it possible that equations with multiple variables have infinitely many roots? It turns out that the latter guess is correct, and in this section you will learn why. 

A variable often denotes an amount. For example, $x $ can denote the amount of apple trees in a garden, and $y $ can denote the amount of cherry trees in the garden. When we solve a problem about the amount of apple trees and cherry trees in a garden, we could get the equation $x = y $. That doesn’t mean that apple trees are cherry trees. Instead, it means that there are as many apple trees as there are cherry trees, since their amount is equal. 

If an equation contains multiple variables, like both $x $ and $y $ for example, an instruction like “solve the equation” is unclear, because we don’t know if we should solve for $x $ or $y $. Therefore, the instruction should either be “solve for $x $” or “solve for $y $”. 

An equation with multiple variables has no specific root. Instead, it can have infinitely many roots. Take the equation $x + y = 1$ for example. If we solve for $x $, we get $x = 1 - y $. The reason why is that if we replace $x $ with $1 - y $ in the original equation, we get $LHS = x + y = 1 - y + y = 1 $ and $RHS = 1 $. Since $LHS = RHS $, $x = 1 - y $ is the root of the equation. The value of $x $ depends on the value of $y $, but $y $ can have infinitely many values, meaning that the same goes for $x $. Therefore, the equation has infinitely many roots. 

Let’s look at the equation $y = x + 1$. That too is an equation with multiple variables, and the value of $y $ depends on the value of $x $ and the other way around. We can try replacing $x $ with different values and see which $y $-values we get. 

$x = -2 $ gives $y = -2 + 1 = -1$.

$x = -1 $ gives $y = -1 + 1 = 0$.

$x = 0 $ gives $y = 0 + 1 = 1$.

$x = 1 $ gives $y = 1 + 1 = 2$.

$x = 2 $ gives $y = 2 + 1 = 3$.

These $5 $ pairs of values can be seen as $5 $ points. We can plot them in a coordinate system. There, the coordinate axes are like number lines, and each point has $2 $ coordinates: an $x $ and a $y $-coordinate. The $x $-coordinate corresponds to the value of $x $, and the $y $ coordinate corresponds to the value of $y $. 

The point $A$ has the coordinates $(-2, -1) $, $B$ has the coordinates$(-1, 0) $, $C$ has the coordinates $(0, 1) $, $D$ has the coordinates $(1, 2) $, and $E$ has the coordinates $(2, 3) $.

We can see that the points form a straight line. So, if we would plot all points with coordinates that solve the equation $y = x + 1 $, we would get a line: 

The line is made up of infinitely many points, and therefore, the equation has infinitely many roots. 

Let’s look at the equation $y = 2x $ instead. It corresponds to a line that looks like this:

If we have two equations, $y = x + 1$ and $y = 2x $, then $x + 1 = 2x $. We can solve this equation: 

$$ x + 1 = 2x $$

$$ x - x + 1 = 2x - x $$

$$ 1 = x $$

$$ x = 1 $$

If we know that $x = 1$, then $y = x + 1$ gives $y = x + 1 = 1 + 1 = 2$.

Now, let’s draw the lines that correspond to the equations $y = x + 1$ and $y = 2x $ in the same coordinate system. 

The lines intersect (meet) in the point A, that has the coordinates $(1, 2) $. Just like the equation $x + 1 = 2x $ has one solution only, the lines intersect in one point only. 

Now, let’s look at the equation $x - 4 = x + 4 $. If we want to do the same thing with this equation, we can draw the lines of the equations $y = x - 4 $ and $y = x + 4 $ in the same coordinate system. 

The lines don’t intersect, which means that the equation $x - 4 = x + 4 $ lacks roots. 

If we look at the equation $10x = 10x $ instead, we can see that we have the same expression on both LHS and RHS. If we would do the same thing with this equation, we would draw the line of the equation $y = 10x $ twice in the same coordinate system: 

Since the lines have the same equation, they coincide with each other, and we can only see one line. All points on one of the lines are points on the other line too, so the lines intersect in infinitely many points. Therefore, the equation $10x = 10x $ has infinitely many roots. 

We can once again go back to the exercise from 1. The equation $252 - u - s = 36 $ may have infinitely many solutions, but we don’t want to know the values of $u $ and $s $. Instead, we want to determine how many coins the three friends have in total. Is there a way to do that?

Earlier, we have seen that equations are statements that describe equalities. We have also mentioned that one side of an equation that stands without the $= $-sign is called an expression. But what is an expression really, and what is the difference between equations and statements? That’s what we are going to find out in this section. 

Let’s look at the expression $6 \cdot 5 + 37 $. It doesn’t contain any variables, and is therefore called a numerical expression. However, the expression $5x + 60 $ contains the variable $x $, and is therefore an algebraic expression. An algebraic expression contains at least one variable, and can also contain coefficients and constants. In the expression $5x + 60 $, $5 $ is a coefficient. It’s a number that a variable is multiplied by. $x $ is a variable term and $60 $ is a constant term. 

Now, let’s look at how an algebraic expression can be created: At a market, bananas and oranges are for sale. The bananas cost $x $ SEK/pc and the oranges cost $y $ SEK/pc. If you buy $15 $ bananas and $7 $oranges, you have to pay $15x $ SEK for the bananas and $7y $ SEK for the oranges. That’s $15x + 7y $ SEK in total. So, $15x + 7y $ is an expression for the sum in SEK that you have to pay for $15 $ bananas and $7 $ oranges. If $15x + 7y $ is connected to another expression with an $= $-sign, like if we write $15x + 7y = 100 $ for example, we get an equation. $15x + 7y = 100 $ can be interpreted as “$15 $ bananas and $7 $ apples cost $100 $ SEK”. So the difference between an expression and an equation is like the difference between saying “$15 $ bananas and $7 $ apples” or “$15 $ bananas and $7 $ apples cost $100 $ SEK”.

Yasmin enters a shop and wants to buy envelopes and paper. The envelopes cost $17 $ SEK/pc and the papers cost $22 $ SEK/pc. Yasmin wonders how much she will have to pay. Help Yasmin by finding an expression for the cost of her purchase. Denote the amount of envelopes that she buys by $e $ and the amount of papers that she buys by $p $.

Let’s go back to the exercise from 1. We have got the equation $252 - u - s = 36 $, that has infinitely many roots. But, since $u $, $s $ and $36 $ are the amount of coins that Ulf, Stuart and Erik had from the beginning, $u + s + 36 $ should be an expression for the amount of coins that they have got in total. 

If we add $u $ and $s $ to both sides of the equation, we get: 

$$ 252 - u + u - s + s = u + s + 36 $$

$$ 252 = u + s + 36 $$.

So, the friends have got $252 $ coins in total. 

Equations can be used to solve many types of math problems that can seem tricky. What should you do if, for example, there are more unknown values than you think you can find? There are several tricks for such problems, and in this section, you will learn these and become a master in problem solving with equations. 

Mia has twice as many apples as Lena. If Mia gives Lena $15 $ apples, they both have the same number of apples. How many apples does Lena have?

I thought of a number. I multiplied it by $3 $ and added $1 $ to the product. Then, I divided this sum by $10 $ and subtracted $5 $ from the quotient. I got $-4 $. Which number did I think of?

When solving some problems, it can seem like there are so many variables that it’s impossible to find all unknown values. In those cases, one variable can often be rewritten so that it only contains another variable. Let’s say that we have got an exercise where we have to determine how many candy bars Peter and Anna have, and we know that Peter has twice as many candy bars as Anna. We can denote the number of candy bars that Peter has by $x $, and the number of candy bars that Anna has by $y $. So, we have $2 $ variables, $x $ and $y $, but we know that $x $ can be written as $2y $. This is called that $x $ is expressed in terms of $y $. If we write $2y $ instead of $x $, we only use the variable $y $. 

If we know something more about $x $ and $y $ and can formulate an equation with these, we can replace $x $ with $2y $. Then, we will get an equation with one variable instead of $2 $, and it’s likely that we will be able to find one solution to that equation. This method is called substitution. Substitute means “replace”, and the idea is to replace a variable with an expression that contains another variable.  

Frida is twice as old as her little brother Adam. Frida is 5 years older than Adam. How old are the two siblings?

Jenny has got twice as many pens as Isabelle, and Isabelle has got twice as many pens as Hedvig. They have got $35 $ pens in total. How many pens have they got each?