Algebraic solution for linear equations with two unknowns

A system of linear equations is, in fact, a set of conditions that must be satisfied by specific unknowns.

If we are given a system of linear equations with two unknowns we must identify the specific $X$ and $Y$ values that satisfy both equations at the same time.
Example of a simple system of equations:

$x+y=5$

$y-x=3$

The solution for a system of equations is based on determining the $X$ and the $Y$ that agree with both the first equation and the second equation.
In this case, the solution to the system of equations is: $y=4,x=1$
By inserting these values we can observe that they actually make both equations true.

There are several ways to solve systems of linear equations with two unknowns, in this article we will focus on the algebraic solution.

It all depends on the equations we are presented with and what we are asked to do.
You might be asked to solve the system of equations with the graphical method, you can do it simply with our guide: Algebraic solution for a system of equations with two unknowns. However, if you have the possibility to choose the type of solution you want, it is more practical to choose the algebraic method.
It is not always simple to plot the equations on a graph and, in fact, the graphical method more often than not requires more time than the algebraic method. Therefore, we recommend that, if you are not required to do so, leave your ruler inside your pencil case and avoid drawing unnecessary planes.

For the purpose of solving systems of equations with two unknowns quickly you will need to know the algebraic method.

As its name would suggest, it is a form that uses algebra: that is, mathematical rules, solving exercises or equations without illustrations.
We will categorise the two following methods used within the algebraic solution:

The Substitution method and The Equation method.

We will begin by expanding on each of the methods as well as tips for you to select the best method according to the system you are presented with.

First step:

An unknown is isolated in one of the equations.

Second step:

The unknown that we isolated is then substituted into the other equation of the system and the value of the unknown is then determined.

Third step:

We insert the value of the unknown that we previously discovered into one equation to determine the value of the other.

In order for you to better understand the substitution method let's take a look at the example.
We assure you that, following this example and a little practice, you will be able to confidently use the substitution method.

Take a look at the following system of equations:

$3y-x=4$

$2x-3y=7$

In this system we are presented with two linear equations with two unknowns $X$ and $Y$.
According to the first step, we must isolate any unknowns from the chosen equation.
Which equation and which unknown should we choose?
The truth is that it doesn't really matter which equation and which unknown you choose, as long as you do it correctly you should obtain the right answer.
To try to avoid mistakes and confusion,

we will select the equation in which the unknowns are either not preceded by any coefficient or have a simple coefficient (such as $1$ or $-1$)
This is in the interest of making it easier for us to extract and isolate.
In our example, in the first equation, the coefficient of $X$ is $1$.
Therefore, we will select this equation and isolate $X$.
After the transposition of members we should obtain the following:

$x=3y-4$

Remember, the transposition of members does not modify the equation itself only its appearance, therefore, the isolated equation and the original one are of the same value.

Let's move on to the second step:

Let us insert the unknown that we have isolated $X$ into the second equation of the system.
Whilst the X has no numerical value, it does have an equivalent expression which is $3y-4$.

Let us take the second equation of the system
$2x-3y=7$

and insert the X $3y-4$ as follows:
$2\times3y-4-3y=7$

Pay attention!

It is very important to add parentheses to the expression in which you have substituted the $X$ in order to avoid confusion.
In this example the coefficient $2$ acts on the whole expression and, if we had forgotten the brackets, we may have thought that we had to multiply the $2$ only by the $3Y$.
Hence do not forget the brackets when substituting an expression in place of the unknown.
In fact, what we did was to change the $X$ in the expression that had only $Y$.
In this way we obtained an equation with only one unknown that is easy for us to solve.
Now let's continue to solve the equation and let's clear the $Y$:

Open the parentheses, insert the elements, transpose the members, and clear the : . $Y$

$6y-8-3y=7$
$3y=15$
$y=5$

Wait a moment, we are not done yet. To solve the system completely we must find the two unknowns.

Let's move on to the third step:

We now know that $Y=5$.
All we have left to do in order to clear the $X$ is to simply insert the $Y$ we found, in our example the $5$, into one of the equations.
Regardless of which equation we choose we should obtain the same $X$.
We recommend that you to go directly to the equation in which we isolated the $X$, insert $Y=5$ and calculate the value of $X$.

Let us return to the isolated equation of $X$:
$x=3y-4$

We already know how much it is worth $Y$is no longer an unknown, so we can write the following:

$y=5$

and we will obtain:

$x=3\times5-4$

$​​​​​​​x=11$

That's it, we are done. We correctly determined the pair of values of $X$ and $Y$ that satisfy the two equations of the system.

Given that we will not always be presented with equations without a coefficient before the unknown or with a single coefficient. It is crucial to remember that when trying achieve isolation we must avoid creating fractions in the equation.

For example, in the system of equations:

$4x+2y=10$

$3x-5y=4$

We must first identify which equation lends itself more to the isolation of unknowns.
Whilst observing the coefficients we notice that in the first equation we will be able to divide the coefficient $2$ by $4$ and by $10$ in integer form, on the other hand, if we try to isolate an unknown in the second equation it will result in an equation with fractions.
Therefore, in this example, if we want to make use of the substitution method, it is preferable to isolate the $Y$ from the first equation.
Consider these questions like you would a puzzle in which you must discover its pieces.
After identifying the first unknown the way to discover the second one is significantly easier.
The key to success in exercises of this type is practice, practice, and practice again.
If you practice the substitution method enough times it should become like second nature to you.

Remember that there are two methods connected to algebraic solving.
The second method is the equating method.

The name of this method is the equalisation method.
All that is required is to equalise the coefficients, in some cases we do not even have to isolate unknowns.
How is it done?

Let's take the following system of equations as an example:
Example of a solution by equating coefficients when the coefficients are equal in one of the unknowns:

$5X+6Y=7$
$5X+4Y=13$

Look at the following equations and notice that the coefficients of the unknown $X$ are identical in both.
In both equations, the coefficient of $X$ is $5$.
When there are equal coefficients in either unknown of the two equations we can subtract the equations from each other.
Simply add the subtract sign as follows:

It is important that you arrange the equations in such a way that the unknowns lie one above the other correspondingly.

$X$ on $X$

$Y$ over $Y$

Number over number.
Once we have the equation $2y=-6$
We can clear the $Y$ and obtain the following $y=-3$

We can then proceed to find the unknown.
Insert X $y=-3$ into one of the equations and we obtain $x=5$.

Regardless of which equation we choose to insert $y=-3$ into we should achieve the same result.
The equalisation method is ideal when you identify identical coefficients of equal unknowns in both equations.

Sometimes the coefficients will not be identical in the two equations.
Even when the coefficients are opposite, i.e., minus and plus, it is still practical to use the equalisation method. However before we can use it we must better understand its meaning.

Let us take the following system of equations as an example:
Example of a solution by equating coefficients when they are opposite in one of the unknowns:

$5X+6Y=7$
$-5X+4Y=13$

In fact, this system of equations is similar to the one we presented in the previous example.
You may notice that the coefficients of the $X$ in this system of equations are not identical. It is true that $5$ and $-5$ are similar, but they are not identical.
If we were to subtract the equations, in the same way as we have done in the previous example, we would obtain a new equation with two unknowns, which would ultimately be futile.
The point of the equalization method is to suppress one variable, to make it disappear completely.

So what can we do to eliminate the unknown $X$ in this system?
Add the equations instead of subtracting them!

By adding the equations we will suppress the coefficient of the $X$ and we will be left with only one unknown.

We add the equations in the same way as we have done in the first example, term over term correspondingly as seen below:

$10Y=20$
We proceed to isolate the variable and obtain the following . $Y$ $y=2$

Do you think we are done solving the system of equations? No, no, no!
To completely solve the system of equations we must identify the $X$ and the $Y$.
Insert the $y=2$ value into the equation you want in order to determine the value of $X$.
Let's insert, for example , $y=2$ into the first equation as follows:

$(5x+6\times2=7$

$5x=-5$
$x=-1$

The result of the system of equations is $x=-1, y=2$

Do you want to make sure you are correct?
Insert into both equations the values you have found and check that the equations remain correct.

What happens when we are given a system of equations with coefficients that are neither equal nor opposite and we are asked to solve it with the equalisation method?
Let's look at an example of this.

Example of a solution by equating coefficients when the coefficients are neither equal nor opposite:

$3X-2Y=11$
$6x+5y=4$

No coefficient of the unknowns is equal or opposite to any other.
Therefore, we must perform a preliminary step before adding or subtracting.
The preliminary step should equal or oppose the coefficient of an unknown, in both equations.

If we carry out a mathematical operation such as division or multiplication, on both sides, the equation may look different, but it will be worth the same as the original, this is the key to the solution of this type of questions.

Look at the system of equations above.
Focus on the coefficients and notice the following eventuality:
If we observe the coefficients of the $X$ we observe that with a simple operation we can convert the $3$ to $6$, simply multiplying by $2$.
and what about the coefficients of the $Y$? Here it will be a little more complicated due to the fact that we will have to perform a mathematical operation on both equations.
Thus, we will convert the coefficient in the first equation to $6$ given that all we will have to perform is a simple mathematical operation: multiply by $2$ only to both members of the first equation.

$3X-2Y=11$

$6x+5y=4$

We will obtain:

$2\times3x-2y=22$

$6x+5y=4$

Remember! To obtain an equivalent equation you must multiply both members of the equation.
If in a certain member there is an expression like the one in our example do not forget to include it in parentheses and multiply everything inside by the desired term.
We will continue solving parentheses and we will obtain:

$6x-4y=22$

This is the new equivalent equation.
We have managed to equalize the coefficients of the $X$ in both equations.
Now, we will apply the method we learned to use when the coefficients are equal in any of the unknowns.
Let's write the equations, one below the other in the correct order, subtract the equations, find an unknown, place it in one of the equations and obtain the second unknown.

Let's not forget to place the $Y$ in one of the original equations to find the $X$, we will obtain the following $x=7/3$.

In fact, when you have a system of linear equations with two unknowns and you want to use the equating method, first look at the equations and observe which case the system corresponds to.
If it corresponds to the case where there are equal coefficients in any unknown, opposite or completely different, then you will be able to choose the most efficient and correct way to solve it.

Sometimes we might encounter a system of equations that has no solution or that has infinite solutions.
If we are given a system of two equations with equal coefficients in two unknowns of the two equations, that is, the coefficient of the $X$ is the same in both equations and also the coefficient of the $Y$ is equal in both equations, but the free number is different, we will obtain the expression of:

$0=$ any number other than $0$

This expression is false!
$0$ cannot be equal to any number other than $0$, therefore, we will say that this system of equations has no solution.

On the other hand, if the coefficients of the two unknowns in both equations are equal and so is the free number (two identical equations), we can immediately determine that this system has infinite solutions.
Why? Because no matter which $X$ and which $Y$ we choose, since we are dealing with two totally identical equations, the expression that will be obtained will always be the same.
Another way to understand this with the coefficient method is to subtract the identical equations and obtain the following $0=0$. True expression.
Great! Now you know how to solve algebraically systems of linear equations with two unknowns.
Wait a minute... But what happens when they don't give you the system and you have to construct it based on a verbal problem?
It's good that you asked.

Sometimes, when we are required to think a little more and to deduce the data on our own, they will not give us a given system of equations. Instead we will be presented with a verbal problem from which we must deduce which equations are appropriate.
In general, we will have to understand the conditions in the problem and based on that create the correct equations.

Here is the problem

What is the price of a pair of pants and the price of a shirt if we know that the pants cost twice as much as the shirt and that the cost of $5$ pants is $22$$ more than the cost of $8$ shirts?

OK.
You might be looking at the problem and wondering how you can transition from it to a system of equations with two unknowns.
Don't worry, it's not that complicated. You just have to concentrate and read the problem carefully.
In the first step we will read the problem without writing the parameters.
At first glance we understand that in this case there are two unknowns that we have to determine, the price of the pants and the price of the shirt.
We are given information about the prices and also certain conditions that must be met, for example, the price of the pants is twice the price of the shirt.
In the second step we will name the unknowns with the letters $X$ and $Y$.
We will mark them and write at random :

Price of the shirt $=X$
Price of the pants. $=Y$

The third step is to transpose the given parameters verbally into the corresponding equations.
How will we do it?
Let's start reading the question again and we soon come across the first condition: the price of the pants is twice the price of the shirt.
In other words, for the price of the pants to be equal to the price of the shirt, the price of the shirt must be multiplied by $2$.
We understand that this might seem somewhat confusing, concentrate and you will see that, since the price of the pants is double that of the shirt, to make them equal we must multiply the price of the shirt in the following way:

$y=2x$

Note that we have assigned $X$ to the price of the shirt and $Y$ to the price of the pants.
This is our first equation in our system of equations.
Continue reading the problem and you will come across the second condition: the cost of $5$ pants is $22$$ greater than the cost of $8$ shirts.
In other words, in order to create an equation, to match the price of a shirt with the price of a pair of pants, we will have to perform several operations.
This condition is a little more complicated than the previous one, however, if you understand the technique, it shouldn't be problematic.
The price of $5$ pants, i.e. $5Y$
is $22$$ more than the price of $8$ shirts, i.e. $8X$.
In fact, we will have to add $22$ to the price of $8$ shirts $=8X$ to equal the cost of $5$ pants $=5Y$.
Let us express this in the equation:

$5y=8x+22$

Another way to understand this condition is to think that the difference between the price of $5$ pants and the price of $8$ shirts is $22$In this way we would obtain an equation equivalent to the one we have found, only in a different order:

$5y-8x=22$

Now we also have the second equation and we can bring it into the system of equations:

$y=2x$
$5y=8x+22$

We can choose the method we want: equalization or substitution and find the two unknowns.
In this case, the $Y$ being already isolated, we would recommend just placing it in the second equation, finding the value of $X$ and then not forgetting to find the value of $Y$ again.

Remember what you had been asked in the problem: What is the price of a pair of pants and the price of a shirt?
In this example:

$5 \times 2x=8x+22$
$10x=8x+22$
$2x=22$
$x=11$

We point to $X$ as the price of the shirt and, therefore, the price of the shirt is $11$$.
Now let's go on to calculate the price of the pants:
Let's put $x=11$

in a very simple equation:
$​​​​​​​y=2x$

and we will obtain:
$y=22$

That is, the price of the pants is $22$$.

The best way to find the answer to this type of problem is to read the question several times and understand exactly what it tells us.
Work through the steps we have detailed above, practice other problems, try different cases and you will undoubtedly master the topic.

If you are interested in this article you may also be interested in the following articles

• Unknowns and Algebraic Expressions
• First-degree equations with an unknown variable
• What is the unknown of a mathematical equation?
• System of Two Linear Equations with Two Unknowns
• Linear equation with two unknowns
• Solution with graphical method for a system of linear equations with two unknowns
• Solving with the substitution method for systems of two linear equations with two unknowns
• Solving by the equating method for systems of two linear equations with two unknowns
• Solving verbal problems with a system of linear equations

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