Equations online. How to solve equations with fractions. Exponential solution of equations with fractions
Fractional equations. ODZ.
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We continue to master the equations. We already know how to work with linear and quadratic equations. The last view left - fractional equations. Or they are also called much more respectably - fractional rational equations. It's the same thing.
Fractional equations.
As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in denominator. At least in one. For example:
Let me remind you that if the denominators are only numbers, these are linear equations.
How to decide fractional equations? First of all, get rid of fractions! After this, the equation most often turns into linear or quadratic. And then we know what to do... In some cases it can turn into an identity, such as 5=5 or an incorrect expression, such as 7=2. But this rarely happens. I will mention this below.
But how to get rid of fractions!? Very simple. Applying the same identical transformations.
We need to multiply the entire equation by the same expression. So that all denominators are reduced! Everything will immediately become easier. Let me explain with an example. Let us need to solve the equation:
How were you taught in elementary school? We move everything to one side, bring it to a common denominator, etc. Forget how bad dream! This is what you need to do when you add or subtract fractions. Or you work with inequalities. And in equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all the denominators (i.e., in essence, by a common denominator). And what is this expression?
On the left side, reducing the denominator requires multiplying by x+2. And on the right, multiplication by 2 is required. This means that the equation must be multiplied by 2(x+2). Multiply:
This is a common multiplication of fractions, but I’ll describe it in detail:
Please note that I am not opening the bracket yet (x + 2)! So, in its entirety, I write it:
On the left side it contracts entirely (x+2), and on the right 2. Which is what was required! After reduction we get linear equation:
And everyone can solve this equation! x = 2.
Let's solve another example, a little more complicated:
If we remember that 3 = 3/1, and 2x = 2x/ 1, we can write:
And again we get rid of what we don’t really like - fractions.
We see that to reduce the denominator with X, we need to multiply the fraction by (x – 2). And a few are not a hindrance to us. Well, let's multiply. All left side and all right side:
Parentheses again (x – 2) I'm not revealing. I work with the bracket as a whole as if it were one number! This must always be done, otherwise nothing will be reduced.
With a feeling of deep satisfaction we reduce (x – 2) and we get an equation without any fractions, with a ruler!
Now let’s open the brackets:
We bring similar ones, move everything to the left side and get:
But before that we will learn to solve other problems. On interest. That's a rake, by the way!
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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 can be 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 area comes into play. acceptable values(ODZ) are such values of the roots of the equation at 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.
Equations with fractions themselves are not difficult and are very interesting. Let's consider the types fractional equations and ways 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. General view such an equation – x/a + b = c, where x is the unknown, a, b and c – regular 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 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. The indication of these conditions is a mandatory and integral part of the right 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 VA.
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 smallest number, which 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 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.