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Solve Equation 5. Solving Linear Equations with Examples

An equation with one unknown, which, after opening the brackets and bringing similar terms, takes the form

ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we’ll figure out how to solve these linear equations.

For example, all equations:

2x + 3= 7 – 0.5x; 0.3x = 0; x/2 + 3 = 1/2 (x – 2) - linear.

The value of the unknown that turns the equation into a true equality is called decision or root of the equation .

For example, if in the equation 3x + 7 = 13 instead of the unknown x we ​​substitute the number 2, we obtain the correct equality 3 2 +7 = 13. This means that the value x = 2 is the solution or root of the equation.

And the value x = 3 does not turn the equation 3x + 7 = 13 into a true equality, since 3 2 +7 ≠ 13. This means that the value x = 3 is not a solution or a root of the equation.

Solution of any linear equations reduces to solving equations of the form

ax + b = 0.

Let's move the free term from the left side of the equation to the right, changing the sign in front of b to the opposite, we get

If a ≠ 0, then x = ‒ b/a .

Example 1. Solve the equation 3x + 2 =11.

Let's move 2 from the left side of the equation to the right, changing the sign in front of 2 to the opposite, we get
3x = 11 – 2.

Let's do the subtraction, then
3x = 9.

To find x, you need to divide the product by a known factor, that is
x = 9:3.

This means that the value x = 3 is the solution or root of the equation.

Answer: x = 3.

If a = 0 and b = 0, then we get the equation 0x = 0. This equation has infinitely many solutions, since when we multiply any number by 0 we get 0, but b is also equal to 0. The solution to this equation is any number.

Example 2. Solve the equation 5(x – 3) + 2 = 3 (x – 4) + 2x ‒ 1.

Let's expand the brackets:
5x – 15 + 2 = 3x – 12 + 2x ‒ 1.


5x – 3x ‒ 2x = – 12 ‒ 1 + 15 ‒ 2.

Here are some similar terms:
0x = 0.

Answer: x - any number.

If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when we multiply any number by 0 we get 0, but b ≠ 0.

Example 3. Solve the equation x + 8 = x + 5.

Let’s group terms containing unknowns on the left side, and free terms on the right side:
x – x = 5 – 8.

Here are some similar terms:
0х = ‒ 3.

Answer: no solutions.

On Figure 1 shows a diagram for solving a linear equation

Let's draw up a general scheme for solving equations with one variable. Let's consider the solution to Example 4.

Example 4. Suppose we need to solve the equation

1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.

2) After reduction we get
4 (x – 4) + 3 2 (x + 1) ‒ 12 = 6 5 (x – 3) + 24x – 2 (11x + 43)

3) To separate terms containing unknown and free terms, open the brackets:
4x – 16 + 6x + 6 – 12 = 30x – 90 + 24x – 22x – 86.

4) Let us group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x – 30x – 24x + 22x = ‒ 90 – 86 + 16 – 6 + 12.

5) Let us present similar terms:
- 22x = - 154.

6) Divide by – 22, We get
x = 7.

As you can see, the root of the equation is seven.

Generally such equations can be solved using the following scheme:

a) bring the equation to its integer form;

b) open the brackets;

c) group the terms containing the unknown in one part of the equation, and the free terms in the other;

d) bring similar members;

e) solve an equation of the form aх = b, which was obtained after bringing similar terms.

However, this scheme is not necessary for every equation. When solving many more simple equations you have to start not from the first, but from the second ( Example. 2), third ( Example. 1, 3) and even from the fifth stage, as in example 5.

Example 5. Solve the equation 2x = 1/4.

Find the unknown x = 1/4: 2,
x = 1/8
.

Let's look at solving some linear equations found in the main state exam.

Example 6. Solve the equation 2 (x + 3) = 5 – 6x.

2x + 6 = 5 – 6x

2x + 6x = 5 – 6

Answer: - 0.125

Example 7. Solve the equation – 6 (5 – 3x) = 8x – 7.

– 30 + 18x = 8x – 7

18x – 8x = – 7 +30

Answer: 2.3

Example 8. Solve the equation

3(3x – 4) = 4 7x + 24

9x – 12 = 28x + 24

9x – 28x = 24 + 12

Example 9. Find f(6) if f (x + 2) = 3 7's

Solution

Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.

We solve the linear equation x + 2 = 6,
we get x = 6 – 2, x = 4.

If x = 4 then
f(6) = 3 7-4 = 3 3 = 27

Answer: 27.

If you still have questions or want to understand solving equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!

TutorOnline also recommends watching a new video lesson from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.

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We solve the fractional rational equation 5/x = 100. This equation can be solved in two ways. Let's look at each of them.

Plan for solving the equation 5/x = 100

  • find the range of acceptable values ​​for a given equation;
  • the first way to solve the equation is by considering it as a proportion;
  • The second way to solve the equation is by finding the unknown divisor.

Finding the unknown term of the proportion

First, let's find the ODZ equation. There is a fraction sign on the left side of the equation and it is equivalent to the division sign. It is known that you cannot divide by zero. This means that from the ODZ we must exclude values ​​that turn the denominator to zero.

ODZ: x belongs to R\(0).

Now let's look at our equation as a proportion.

The main property of proportion.

The product of the extreme terms of a proportion is equal to the product of its middle terms.

For proportion a: b = c: d or a/b = c/d the main property is written like this: a · d = b · c.

Let's apply it and get a linear equation:

100 * x = 5 * 1;

Let's divide both sides of the equation by 100, thereby getting rid of the coefficient in front of the x variable:

Finding an unknown divisor

Let's look at the equation as a quotient. Where the dividend is 5, the divisor is x, and the result of division is the quotient is 100.

Let's remember the rule for finding an unknown divisor - you need to divide the dividend by the quotient.

The found root belongs to the ODZ equation.

Let's check the found solution to the equation. To do this, substitute the found roots into the original equation and perform the calculations:

The solution was found correctly.

An equation is an equality in which there is an unknown term - x. Its meaning must be found.

The unknown quantity is called the root of the equation. Solving an equation means finding its root, and to do this you need to know the properties of the equations. The equations for grade 5 are not difficult, but if you learn to solve them correctly, you will not have problems with them in the future.

The main property of the equations

When both sides of an equation change by the same amount, it continues to be the same equation with the same root. Let's solve some examples to better understand this rule.

How to Solve Equations: Addition or Subtraction

Suppose we have an equation of the form:

  • a + x = b - here a and b are numbers, and x is the unknown term of the equation.

If we add (or subtract from them) the value c to both sides of the equation, it will not change:

  • a + x + c = b + c
  • a + x - c = b - c.

Example 1

Let's use this property to solve the equation:

  • 37+x=51

Subtract the number 37 from both sides:

  • 37+x-37=51-37

we get:

  • x=51-37.

The root of the equation is x=14.

If we look closely at the last equation, we can see that it is the same as the first one. We simply moved term 37 from one side of the equation to the other, replacing plus with minus.

It turns out that any number can be transferred from one part of the equation to another with the opposite sign.

Example 2

  • 37+x=37+22

Let's carry out the same action, move the number 37 from the left side of the equation to the right:

  • x=37-37+22

Since 37-37=0, we simply reduce this and get:

  • x =22.

Identical terms of an equation with the same sign, located in different parts of the equation, can be reduced (crossed out).

Multiplying and dividing equations

Both sides of the equality can also be multiplied or divided by the same number:

If the equality a = b is divided or multiplied by c, it does not change:

  • a/c = b/c,
  • ac = bс.

Example 3

  • 5x = 20

Let's divide both sides of the equation by 5:

  • 5x/5 = 20/5.

Since 5/5 = 1, we reduce these multiplier and divisor on the left side of the equation and get:

  • x = 20/5, x=4

Example 4

  • 5x = 5a

If both sides of the equation are divided by 5, we get:

  • 5x/5 = 5a/5.

The 5 in the numerator and denominator of the left and right sides are canceled, resulting in x = a. This means that identical factors on the left and right sides of the equations cancel.

Let's solve another example:

  • 13 + 2x = 21

We move term 13 from the left side of the equation to the right with the opposite sign:

  • 2x = 21 - 13
  • 2x = 8.

Dividing both sides of the equation by 2, we get:

  • x = 4.

One of the most important skills when admission to 5th grade is the ability to solve simple equations. Since 5th grade is not yet so far from primary school, then there are not so many types of equations that a student can solve. We will introduce you to all the basic types of equations that you need to be able to solve if you want enter a physics and mathematics school.

Type 1: "bulbous"
These are equations that you are almost likely to encounter when admission to any school or 5th grade club as separate task. They are easy to distinguish from others: in them the variable is present only once. For example, or.
They are solved very simply: you just need to “get” to the unknown, gradually “removing” everything unnecessary that surrounds it - as if peeling an onion - hence the name. To solve it, just remember a few rules from the second class. Let's list them all:

Addition

  1. term1 + term2 = sum
  2. term1 = sum - term2
  3. term2 = sum - term1

Subtraction

  1. minuend - subtrahend = difference
  2. minuend = subtrahend + difference
  3. subtrahend = minuend - difference

Multiplication

  1. factor1 * factor2 = product
  2. factor1 = product: factor2
  3. factor2 = product: factor1

Division

  1. dividend: divisor = quotient
  2. dividend = divisor * quotient
  3. divisor = dividend: quotient

Let's look at an example of how to apply these rules.

Note that we are dividing on and we receive . In this situation, we know the divisor and the quotient. To find the dividend, you need to multiply the divisor by the quotient:

We have become a little closer to ourselves. Now we see that is added and it turns out . This means that to find one of the terms, you need to subtract the known term from the sum:

And another “layer” has been removed from the unknown! Now we see a situation with a known value of the product () and one known factor ().

Now the situation is “minuend - subtrahend = difference”

And the last step - famous work() and one of the multipliers ()

Type 2: equations with brackets
Equations of this type are most often found in problems - 90% of all problems for admission to 5th grade. Unlike "onion equations" the variable here can appear several times, so it is impossible to solve it using the methods from the previous paragraph. Typical equations: or
The main difficulty is opening the brackets correctly. After you have managed to do this correctly, you should reduce similar terms (numbers to numbers, variables to variables), and after that we get the simplest "onion equation" which we can solve. But first things first.

Expanding parentheses. We will give several rules that should be used in this case. But, as practice shows, the student begins to open the brackets correctly only after 70-80 completed problems. The basic rule is this: any factor outside the brackets must be multiplied by each term inside the brackets. And the minus sign in front of the bracket changes the sign of all the expressions inside. So, the basic rules of disclosure:










Bringing similar. Here everything is much easier: you need, by transferring the terms through the equal sign, to ensure that on one side there are only terms with the unknown, and on the other - only numbers. The basic rule is this: each term transferred through changes its sign - if it was with, it will become with, and vice versa. After a successful transfer, it is necessary to count the total number of unknowns, the total number on the other side of the equality than the variables, and solve a simple "onion equation".

Application

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