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 find the specific $X$ and $Y$ 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 finding 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 placing these values we will see 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 generally convenient to choose the algebraic method.
It is not always easy to plot the equations on a graph and, in fact, the graphical method sometimes takes longer than the algebraic method. Therefore, we recommend that, if you are not required to do so, you leave the ruler inside the case and avoid drawing unnecessary planes.

To solve systems of equations with two unknowns quickly you will need to know the algebraic method.

As its name says, it is a form that uses algebra: that is, mathematical rules, solving exercises or equations without illustrations.
We will distinguish two methods within the algebraic resolution:

Substitution method and Equation method.

We will explain about each of them and give you tips to choose the best method according to the system you are presented with.

First step:

An unknown is isolated in any of the equations.

Second step:

The unknown we isolated is substituted into the other equation of the system and the value of an unknown is discovered.

Third step:

We place the value of the unknown we discovered in one equation to find the value of the other.

In order for you to understand the substitution method we will see it in an example.
We assure you that, after the example and a little practice, you will be able to use the substitution method without any problem.

Let's take for example the following system of equations:

$3y-x=4$

$2x-3y=7$

In this system there are two linear equations with two unknowns $X$ and $Y$.
According to the first step, we must isolate any unknowns of the equation we choose.
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 will arrive at the right answer.
To try to avoid mistakes and confusion,

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

$x=3y-4$

Remember, the transposition of members and the rescue do not modify the equation itself, but only its appearance, therefore, the isolated equation and the original one are worth the same.

Let's move on to the second step:

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

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

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

Attention!

It is very important to add parentheses to the expression in which you have substituted the $X$ so that you do not get confused.
In this example the coefficient $2$ acts on the whole expression and, if we had forgotten the brackets, we could have thought that we had to multiply the $2$ only by the $3Y$.
So, do yourself a favor and do not forget the brackets when substituting some 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 with solving the equation and let's clear the $Y$:

Let's open parentheses, place elements, transpose 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 to clear the $X$ is to simply put the $Y$ we found, in our example the $5$, into one of the equations.
Regardless of which equation we choose we will get the same $X$.
We recommend you to go directly to the equation in which we isolated the $X$, put $Y=5$ and discover very easily 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 will write:

$y=5$

and we will get:

$x=3\times5-4$

$​​​​​​​x=11$

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

We will not always have equations with no coefficient before the unknown or with a single coefficient. What we must take into account to achieve isolation in the simplest way is to avoid that such isolation creates fractions in the equation.

For example, in the system of equations:

$4x+2y=10$

$3x-5y=4$

We will ask ourselves in which equation it will be easier to isolate.
Taking a look at the coefficients we will 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 some unknown in the second equation we will have an equation with fractions.
Therefore, in this example, if we want to make use of the substitution method, we will prefer to isolate the $Y$ from the first equation.
Consider these questions as if it were a puzzle in which you must discover its pieces.
After clearing the first unknown the way to discover the second one is easy and fast.
The key to success in exercises of this type is practice, practice, and practice again.
If you practice it, you will completely assimilate the substitution method and you will be able to use it naturally.

Remember we had said that there are two methods in algebraic solving?
The second method is the equating method.

The name of this method is the method of equalization.
All we have to do is to equalize 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
in fact, we have obtained the equation $2y=-6$
we will clear the $Y$ and we will obtain $y=-3$

Now we will find the unknown.
Put X $y=-3$ into one of the equations and we get $x=5$.

Regardless of which equation we choose to put $y=-3$ in we will arrive at the same result.
When you identify identical coefficients of equal unknowns in both equations, you will want to use the equating method.

Sometimes the coefficients will not be identical in the two equations.
Even when the coefficients are opposite, i.e., minus and plus, it is convenient to use the equalization method. 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 gave in the previous example.
Detailers will 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 arrive at a new equation with two unknowns, which would not really lead us anywhere.
The point of the equalization method is to get 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 and we get:

$10Y=20$
we will isolate the and get . $Y$ $y=2$

Do you think we are done solving the system of equations? No, no, no!
To completely solve the system of equations we cannot forget to find the $X$ and the $Y$.
Put the $y=2$ you found in the equation you want and find out the value of $X$.
Let's put, for example , $y=2$ in the first equation and we will get:

$(5x+6\times2=7$

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

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

Want to make sure you got it right?
Put in 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 equating 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 do a preliminary step before adding or subtracting.
The preliminary step is to equal or oppose the coefficient of an unknown, in both equations.

If we carry out some 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 look at the coefficients of the $X$ we will see 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 because we will have to perform a mathematical operation on the two equations.
So, we will decide that we want to convert the coefficient in the first equation to $6$ because all we will have to do 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 arrive at 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 get the second unknown.

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

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 see 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 get $0=0$. 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?
Lucky you ask.

Sometimes, when they want us to think a little more and to deduce the data by our own means, they will not give us a given system of equations but 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 get 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 here there are two unknowns that we have to find out, 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 and write randomly:

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 will 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 a bit 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 first noted $X$ as the price of the shirt and $Y$ as the price of the pants.
This is our first equation in our system of equations.
Now we will continue reading the problem and come across the second condition: the cost of $5$ pants is $22$$ greater than the cost of $8$ shirts.
In other words, to create an equation, to match the price of a shirt with the price of a pair of pants, we will have to do several operations.
This condition is a little more complicated than the previous one, but, if you understand the technique, it will come out easily.
The price of $5$ pants, i.e. $5Y$
is $22$$ 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 be sure to master the topic exceptionally well.

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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