Solving equations with different denominators. Equations online

Equations with fractions themselves are not difficult and are very interesting. Let's look at the types of fractional equations and how to solve them.

How to solve equations with fractions - x in the numerator

If a fractional equation is given, where the unknown is in the numerator, the solution does not require additional conditions and is solved without unnecessary hassle. The general form of such an equation is x/a + b = c, where x is the unknown, a, b and c are ordinary numbers.

Find x: x/5 + 10 = 70.

In order to solve the equation, you need to get rid of fractions. Multiply each term in the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 are cancelled, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 – 50 = 300.

Find x: x/5 + x/10 = 90.

This example is a slightly more complicated version of the first one. There are two possible solutions here.

Option 1: We get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90×10 => 2x + x = 900 => 3x = 900 => x=300.
Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. Divide 10 by 10, multiply by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.

We often encounter fractional equations in which the x's are on opposite sides of the equal sign. In such situations, it is necessary to move all the fractions with X's to one side, and the numbers to the other.

Find x: 3x/5 = 130 – 2x/5.
Move 2x/5 to the right with the opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
We reduce 5x/5 and get: x = 130.

How to solve an equation with fractions - x in the denominator

This type of fractional equations requires writing additional conditions. Specifying these conditions is a mandatory and integral part of a correct decision. By not adding them, you run the risk, since the answer (even if it is correct) may simply not be counted.

The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is the unknown, a, b, c are ordinary numbers. Please note that x may not be any number. For example, x cannot equal zero, since it cannot be divided by 0. This is precisely the additional condition that we must specify. This is called the range of permissible values, abbreviated as OA.

Find x: 15/x + 18 = 21.

We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation according to the standard scheme, getting rid of fractions. Multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.

Often there are equations where the denominator contains not only x, but also some other operation with it, for example, addition or subtraction.

Find x: 15/(x-3) + 18 = 21.

We already know that the denominator cannot be equal to zero, which means x-3 ≠ 0. We move -3 to the right side, changing the “-” sign to “+” and we get that x ≠ 3. The ODZ is indicated.

We solve the equation, multiply everything by x-3: 15 + 18×(x – 3) = 21×(x – 3) => 15 + 18x – 54 = 21x – 63.

Move the X's to the right, numbers to the left: 24 = 3x => x = 8.

The lowest common denominator is used to simplify this equation. This method is used when you cannot write a given equation with one rational expression on each side of the equation (and use the crisscross method of multiplication). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, it is better to use criss-cross multiplication).

Find the lowest common denominator of the fractions (or least common multiple). NOZ is the smallest number that is evenly divisible by each denominator.

Sometimes NPD is an obvious number. For example, if given the equation: x/3 + 1/2 = (3x +1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 is 6.
If the NCD is not obvious, write down the multiples of the largest denominator and find among them one that will be a multiple of the other denominators. Often the NOD can be found by simply multiplying two denominators. For example, if the equation is given x/8 + 2/6 = (x - 3)/9, then NOS = 8*9 = 72.
If one or more denominators contain a variable, the process becomes somewhat more complicated (but not impossible). In this case, the NOC is an expression (containing a variable) that is divided by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divided by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).

Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOC by the corresponding denominator of each fraction. Since you are multiplying both the numerator and denominator by the same number, you are effectively multiplying the fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).

So in our example, multiply x/3 by 2/2 to get 2x/6, and 1/2 multiply by 3/3 to get 3/6 (the fraction 3x +1/6 does not need to be multiplied because it the denominator is 6).
Proceed similarly when the variable is in the denominator. In our second example, NOZ = 3x(x-1), so multiply 5/(x-1) by (3x)/(3x) to get 5(3x)/(3x)(x-1); 1/x multiplied by 3(x-1)/3(x-1) and you get 3(x-1)/3x(x-1); 2/(3x) multiplied by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).

Find x. Now that you have reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by the common denominator. Then solve the resulting equation, that is, find “x”. To do this, isolate the variable on one side of the equation.

In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with the same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by N3, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.

Solving equations with fractions Let's look at examples. The examples are simple and illustrative. With their help, you will be able to understand in the most understandable way.
For example, you need to solve the simple equation x/b + c = d.

An equation of this type is called linear, because The denominator contains only numbers.

The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side cancels.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides by 5. We get:
x+20=45
x=45-20=25

Another example when the unknown is in the denominator:

Equations of this type are called fractional-rational or simply fractional.

We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic equation, which is solved in the usual way. You just need to consider the following points:

the value of a variable that turns the denominator to 0 cannot be a root;
You cannot divide or multiply an equation by the expression =0.

This is where the concept of the region of permissible values (ADV) comes into force - these are the values of the roots of the equation for which the equation makes sense.

Thus, when solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.

For example, you need to solve a fractional equation:

Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x – any value other than zero.

We get rid of the denominator by multiplying all terms of the equation by x

And we solve the usual equation

5x – 2x = 1
3x = 1
x = 1/3

Answer: x = 1/3

Let's solve a more complicated equation:

ODZ is also present here: x -2.

When solving this equation, we will not move everything to one side and bring the fractions to a common denominator. We will immediately multiply both sides of the equation by an expression that will cancel out all the denominators at once.

To reduce the denominators, you need to multiply the left side by x+2, and the right side by 2. This means that both sides of the equation must be multiplied by 2(x+2):

This is the most common multiplication of fractions, which we have already discussed above.

Let's write the same equation, but slightly differently

The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain the usual linear equation:

x = 4 – 2 = 2, which corresponds to our ODZ

Answer: x = 2.

Solving equations with fractions not as difficult as it might seem. In this article we have shown this with examples. If you have any difficulties with how to solve equations with fractions, then unsubscribe in the comments.

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. In 5th grade, mathematics students study quite a lot of new topics, one of which will be fractional equations. For many, this is a rather complex topic that parents should help their children understand, and if parents have forgotten mathematics, then they can always use online programs that solve equations. So, using an example, you can quickly understand the algorithm for solving equations with fractions and help your child.

Below, for clarity, we will solve a simple fractional linear equation of the following form:

\[\frac(x-2)(3) - \frac(3x)(2)=5\]

To solve this type of equation, it is necessary to determine the NOS and multiply the left and right sides of the equation by it:

\[\frac (x-2)(3) - \frac(3x)(2)=5\]

This gives us a simple linear equation because the common denominator as well as the denominator of each fractional term cancels out:

Let's move the terms with the unknown to the left:

Let's divide the left and right sides by -7:

From the obtained result we can select an entire part, which will be the final result of solving this fractional equation:

Where can I solve equations with fractions online?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

An equation is an equality containing a letter whose value must be found.

In equations, the unknown is usually represented by a lowercase letter. The most commonly used letters are “x” [ix] and “y” [y].

Root of the equation- this is the value of the letter at which the correct numerical equality is obtained from the equation.

Solve the equation- means to find all its roots or make sure that there are no roots.

Having solved the equation, we always write down a check after the answer.

Information for parents

Dear parents, we draw your attention to the fact that in elementary school and in the 5th grade, children DO NOT know the topic “Negative numbers”.

Therefore, they must solve equations using only the properties of addition, subtraction, multiplication, and division. Methods for solving equations for grade 5 are given below.

Do not try to explain the solution of equations by transferring numbers and letters from one part of the equation to another with a change in sign.

You can brush up on concepts related to addition, subtraction, multiplication and division in the lesson “Laws of Arithmetic”.

Solving addition and subtraction equations

How to find the unknown
term

How to find the unknown
minuend

How to find the unknown
subtrahend

To find the unknown term, you need to subtract the known term from the sum.

To find the unknown minuend, you need to add the subtrahend to the difference.

To find the unknown subtrahend, you need to subtract the difference from the minuend.

x + 9 = 15
x = 15 − 9
x=6
Examination

x − 14 = 2
x = 14 + 2
x = 16
Examination

16 − 2 = 14
14 = 14

5 − x = 3
x = 5 − 3
x = 2
Examination

Solving multiplication and division equations

How to find an unknown
factor

How to find the unknown
dividend

How to find an unknown
divider

To find an unknown factor, you need to divide the product by the known factor.

To find the unknown dividend, you need to multiply the quotient by the divisor.

To find an unknown divisor, you need to divide the dividend by the quotient.

y 4 = 12
y=12:4
y=3
Examination

y: 7 = 2
y = 2 7
y=14
Examination

8:y=4
y=8:4
y=2
Examination

An equation is an equality containing a letter whose sign must be found. The solution to an equation is the set of letter values that turns the equation into a true equality:

Recall that to solve equation you need to transfer the terms with the unknown to one part of the equality, and the numerical terms to the other, bring similar ones and get the following equality:

From the last equality we determine the unknown according to the rule: “one of the factors is equal to the quotient divided by the second factor.”

Since rational numbers a and b can have the same or different signs, the sign of the unknown is determined by the rules for dividing rational numbers.

Procedure for solving linear equations

The linear equation must be simplified by opening the brackets and performing the second step operations (multiplication and division).

Move the unknowns to one side of the equal sign, and the numbers to the other side of the equal sign, obtaining an equality identical to the given one,

Bring similar ones to the left and right of the equal sign, obtaining an equality of the form ax = b.

Calculate the root of the equation (find the unknown X from equality x = b : a),

Check by substituting the unknown into the given equation.

If we obtain an identity in a numerical equality, then the equation is solved correctly.

Special cases of solving equations

If equation given a product equal to 0, then to solve it we use the property of multiplication: “the product is equal to zero if one of the factors or both factors are equal to zero.”

27 (x - 3) = 0
27 is not equal to 0, which means x - 3 = 0

The second example has two solutions to the equation, since
this is a second degree equation:

If the coefficients of the equation are ordinary fractions, then first of all you need to get rid of the denominators. To do this:

Find the common denominator;

Determine additional factors for each term of the equation;

Multiply the numerators of fractions and integers by additional factors and write all terms of the equation without denominators (the common denominator can be discarded);

Move the terms with unknowns to one side of the equation, and the numerical terms to the other from the equal sign, obtaining an equivalent equality;

Bring similar members;

Basic properties of equations

In any part of the equation, you can add similar terms or open a parenthesis.

Any term of the equation can be transferred from one part of the equation to another by changing its sign to the opposite.

Both sides of the equation can be multiplied (divided) by the same number, except 0.

In the example above, all its properties were used to solve the equation.

How to solve an equation with an unknown in a fraction

Sometimes linear equations take the form when unknown appears in the numerator of one or more fractions. Like in the equation below.

In such cases, such equations can be solved in two ways.

I solution method
Reducing an equation to a proportion

When solving equations using the proportion method, you must perform the following steps:

bring all the fractions to a common denominator and add them as algebraic fractions (only one fraction should remain on the left and right sides);

Solve the resulting equation using the rule of proportion.

So let's go back to our equation. On the left side we already have only one fraction, so no transformations are needed in it.

We will work with the right side of the equation. Let's simplify the right side of the equation so that there is only one fraction left. To do this, remember the rules for adding a number with an algebraic fraction.

Now we use the rule of proportion and solve the equation to the end.

II method of solution
Reduction to a linear equation without fractions

Let's look at the equation above again and solve it in a different way.

We see that there are two fractions in the equation "

How to solve equations with fractions. Exponential solution of equations with fractions.

An equation of this type is called linear, because The denominator contains only numbers.

The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side cancels.

For example, how to solve a fractional equation:
x/5+4=9
We multiply both sides by 5. We get:
x+20=45

Another example when the unknown is in the denominator:

Equations of this type are called fractional-rational or simply fractional.

the value of a variable that turns the denominator to 0 cannot be a root;
You cannot divide or multiply an equation by the expression =0.

This is where the concept of the region of permissible values (ADV) comes into force - these are the values of the roots of the equation for which the equation makes sense.

Thus, when solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our ODZ are excluded from the answer.

For example, you need to solve a fractional equation:

Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x – any value other than zero.

We get rid of the denominator by multiplying all terms of the equation by x

And we solve the usual equation

5x – 2x = 1
3x = 1
x = 1/3

Let's solve a more complicated equation:

ODZ is also present here: x -2.

To reduce the denominators, you need to multiply the left side by x+2, and the right side by 2. This means that both sides of the equation must be multiplied by 2(x+2):

This is the most common multiplication of fractions, which we have already discussed above.

Let's write the same equation, but slightly differently

The left side is reduced by (x+2), and the right by 2. After the reduction, we obtain the usual linear equation:

x = 4 – 2 = 2, which corresponds to our ODZ

Solving equations with fractions grade 5

Solving equations with fractions. Solving fraction problems.

View document contents
“Solving equations with fractions, grade 5”

— Addition of fractions with the same denominators.

— Subtraction of fractions with the same denominators.

Adding fractions with like denominators.

To add fractions with the same denominators, you need to add their numerators and leave the denominator the same.

Subtracting fractions with like denominators.

To subtract fractions with the same denominators, you need to subtract the numerator of the minuend from the numerator of the minuend, but leave the denominator the same.