Equation with variable in the denominator

Equation with unknown in denominator

When we have an exercise with the unknown in the denominator it means that there is a solution set, that is to say that there are numbers that if we place them in place of the X X we will obtain a meaningless expression.

When we go to solve an exercise of this type, first of all, we will find out which is the solution set, that is, which number would zero the denominator and would cause the expression to be meaningless.

For example, in the fiscal year

3X=6 \frac{3}{X}=6

The denominator is X X and, therefore, the solution set will be:

X0X ≠0

since if we were to place 0=X 0=X we would receive a meaningless expression (denominator 0 0 ).
After solving the exercise we will verify that the result does not appear in the solution set.

3=6×X 3=6 \times X

3=6X3=6X

X=12X=\frac{1}{2}

Now, we must check that the solution belongs to the set. In our case
X=12X=\frac{1}{2} indeed belongs to the solution set which is X0X ≠0

therefore, X=12X=\frac{1}{2} is the solution.

A - Equation with variable in the denominator

In this article we will learn how to solve equations with an unknown in the denominator. We will start with a simple exercise and then move on.

Examples with solutions of equations with variable in the denominator

Example 1

3X=6 \frac{3}{X}=6

So far we have not performed exercises with a variable in the denominator. In this case, the denominator of the left-hand member of the equation is X X . How does this affect our exercise? This means that the exercise has a solution set (sometimes called a solution set), i.e. there are numbers that if we put them in place of the X X we will get a meaningless expression.

Whenever we go to solve an exercise with an unknown in the denominator, we will first find out what the solution set is, i.e., what number would set the denominator to zero and cause the expression to be meaningless.

In our case, the denominator is X X , therefore, the solution set is

X0X ≠0

  • well, the solution set is any X X as long as
    X0X≠0

    because if we put X0X ≠0 we will get an invalid expression (denominator 0 0 ).

Now we will continue with the exercise. We will want to get rid of the denominator, so we will multiply both sides of the equation by X X .

3=6×X 3=6\times X

and we will get

3=6X3=6X

X=12X=\frac{1}{2}

Now, we must check that the solution belongs to the set. In our case

X=12X=\frac{1}{2}

it does indeed belong to the solution set which is

X0X ≠0

therefore, X=12X=\frac{1}{2}

is the solution.

At this stage it is highly recommended to check it in the following way. We will put X=12X=\frac{1}{2}

in the original expression and verify it:

in the original expression and verify it

That is, the answer is correct.


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Example 2

2X1=3\frac{2}{X-1}=3

First, we will write down the solution set. We must verify that the denominator differs from 0, therefore, we will write down that the solution set is:

X10X-1 ≠0

X1X ≠1

We advise to write it down clearly and prominently. We will return to this at the end of the exercise.

In the meantime we will write down:

Solution set: X1X ≠1

Now we will return to the solution set of the exercise. We will want to get rid of the denominator, therefore, we will multiply the two sides of the equation by X1X-1

We will obtain:
(2X1=3)×(X1)(\frac{2}{X-1}=3)\times (X-1)

2=3×(X1)2=3\times (X-1)

2=3X32=3X-3

5=3X5=3X

X=53X=\frac{5}{3}

Now we will check if the solution is in the set. Recall that the solution set is

X1X ≠1

i.e., the solution we have obtained is in it, therefore, the result is X=53X=\frac{5}{3}

At this stage it is strongly recommended to place the result in the original expression to see if we got it right. Try it!


Example 3

2X33=2\frac{2}{X-3}-3=2

First, we will find out what the solution set is.

X30X-3 ≠0

X3X ≠3

Now we go back to the exercise. We pass the 3 to the other side and continue solving. We will get

2X3=5\frac{2}{X-3}=5

Now we want to get rid of the denominator. We'll multiply both sides of the equation by the denominator and we'll get:

(2X3=5)×(X3)(\frac{2}{X-3}=5)\times (X-3)

2=5×(x3)2=5\times (x-3)

2=5X152=5X-15

5X=175X=17

X=175X=\frac{17}{5}

This solution effectively lies within the solution set which is. X3X ≠3

It is recommended to verify that the result is correct by placing it in the original expression.


Example 4: Equation with more than one denominator

2X3=1X\frac{2}{X-3}=\frac{1}{X}

We will notice that in this exercise there is more than one denominator. First, we will want to find out what the solution set is. We will find out separately for each denominator.

That is, we must first check that:

X30X-3 ≠0
X3X ≠3
In addition, we have to check the denominator located on the right side of the equation, that is:

X0X≠0

Before continuing with the resolution of the exercise we will write it down in a clear and orderly manner.
The solution set is
X3X ≠3
X0X ≠0

Now we will want to get rid of both denominators. To do so, we will multiply both members of the equation by the common denominator. The common denominator in this case is X×(X3)X\times (X-3)

(2X3=1X)×X(X3)(\frac{2}{X-3}=\frac{1}{X})\times X(X-3)

2X3×X(X3)=1X×X(X3)\frac{2}{X-3}\times X(X-3)=\frac{1}{X}\times X(X-3)

2X=X32X=X-3
X=3X=-3

The result X=3X=-3 is compatible with the solution set.

To corroborate our result, we will place it in the original equation and examine if a true equality is obtained:

1(3)=2(33)\frac{1}{(-3)}=\frac{2}{(-3-3)}

1(3)=1(3)\frac{1}{(-3)}=\frac{1}{(-3)}

Then, our result is correct.


Example 5

34X+2=18X\frac{3}{4X}+2=\frac{1}{8X}

First we will find out which is the solution set. We must verify that both denominators differ from zero. In this case the number 0 will set both denominators to zero, so we will write down:

The solution set is X0 X\ne0
Also in this exercise we have two denominators with unknown and we will want to get rid of both. To do this we will multiply them by the common denominator. We will notice that if we multiply 4X4X by 22 we will get 8X8X.
Therefore, the common denominator is 8X8X

(34X+2=18X)×8X(\frac{3}{4X}+2=\frac{1}{8X})\times 8X

34X×8X+2×8X=18X×8X \frac{3}{4X}\times 8X+2\times 8X=\frac{1}{8X}\times 8X

6+16X=16+16X=1
16X=516X=-5

X=516X=-\frac{5}{16}

The result we obtained fits the solution set.
It is highly recommended to place it in the original exercise to corroborate that a true equality is obtained.

34×(516)+2=18×(516)\frac{3}{4\times (-\frac{5}{16})}+2=\frac{1}{8\times (-\frac{5}{16})}

3(54)+2=1(52)\frac{3}{(-\frac{5}{4})}+2=\frac{1}{(-\frac{5}{2})}

125+2=25-\frac{12}{5}+2=-\frac{2}{5}

1210=212-10=2

2=22=2

We have obtained a true equality, therefore, our solution is correct.


Example 6: Removing the common term from the denominator

33X2+1=16X4 \frac{3}{3X-2}+1=\frac{1}{6X-4}
We will notice that the denominator on the right side can be divided into other terms by taking out the common term.

We will do it and we will obtain:

33X2+1=12(3X2) \frac{3}{3X-2}+1=\frac{1}{2(3X-2)}

Now we will continue with the exercise. First we will find the solution set. We will write down

3X203X-2 ≠0

3X23X ≠2

X23X ≠\frac{2}{3}

Let's go back to the exercise. To get rid of the denominator we have to multiply both members of the equation by the common denominator. As we can see, after taking out the common term, it gives us that the common denominator is 2×(3X2)2\times (3X-2)

Therefore, we will multiply both members of the equation by it and we will get

(33X2+1=12(3X2))×2(3X2) (\frac{3}{3X-2}+1=\frac{1}{2(3X-2)})\times 2(3X-2)

33X2×2×(3X2)+1×2×(3X2)=12×(3X2)×2×(3X2) \frac{3}{3X-2}\times 2\times(3X-2)+1 \times 2 \times(3X-2)=\frac{1}{2\times(3X-2)}\times 2\times(3X-2)

6+6X4=16+6X-4 =1

6X=16X=-1

X=16X=-\frac{1}{6}

We will see that the result obtained is indeed in the solution set. X23X ≠\frac{2}{3}

At this stage it is strongly recommended to place the result obtained in the original equation to check if we have done it right. Try it!