One of the most important skills in admission to 5th grade is the ability to solve simple equations. Since the 5th grade is not so far from elementary school, then there are not so many types of equations that a student can solve. We will introduce you to all the main types of equations that you need to be able to solve if you want enroll in a physics and mathematics school.
1 type: "bulbous"
These are equations that you will almost certainly encounter when admission to any school or a 5th grade circle as a separate task. They are easy to distinguish from others: they contain a variable only once. For example, or.
They are solved very simply: you just need to "get" to the unknown, gradually "removing" everything superfluous that surrounds it - as if peeling an onion - hence the name. To solve it, it is enough to remember a few rules from the second class. Let's list them all:
Addition
- term1 + term2 = sum
- term1 = sum - term2
- term2 = sum - term1
Subtraction
- minuend - subtrahend = difference
- minuend = subtrahend + difference
- subtrahend = minuend - difference
Multiplication
- multiplier1 * multiplier2 = product
- multiplier1 = product: multiplier2
- multiplier2 = product: multiplier1
Division
- dividend: divisor = quotient
- dividend = divisor * quotient
- divisor = dividend: quotient
Let's look at an example of how to apply these rules.
Note that we share on and we get . In this situation, we know the divisor and the quotient. To find the dividend, you need to multiply the divisor by the quotient:
We got a little closer to ourselves. Now we see that to added and obtained. So, to find one of the terms, you need to subtract the known term from the sum:
And one more "layer" is removed from the unknown! Now we see the situation with known value products () and one known multiplier ().
Now the situation is "reduced - subtracted = difference"
And the last step is the known product () and one of the factors ()
2 type: equations with brackets
Equations of this type are most often found in problems - 90% of all problems for admission to grade 5. Unlike "onion equations" the variable here can occur several times, so it is impossible to solve it using the methods from the previous paragraph. Typical equations: or
The main difficulty is to correctly open the brackets. After we managed to do this correctly, we should bring like terms (numbers to numbers, variables to variables), and after that we get the simplest "onion equation" which we can solve. But first things first.
Bracket expansion. We will give a few rules that should be used in this case. But, as practice shows, the student begins to correctly open the brackets only after 70-80 solved problems. The basic rule is this: any factor outside the brackets must be multiplied by each term inside the brackets. And the minus before the bracket changes the sign of all expressions that are inside. So, the basic rules of disclosure:
Bringing similar. Everything is much easier here: by transferring the terms through the equal sign, you need to ensure that on the one hand there are only terms with the unknown, and on the other - only numbers. The basic rule is this: each term carried through changes its sign - if it was with, then it will become with, and vice versa. After a successful transfer, it is necessary to count the total number of unknowns, the final number on the other side of equality than the variables, and solve a simple "onion equation".
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 equation 5/x = 100
- find the range of admissible values for the given equation;
- the first way to solve an 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. We know that you can't divide by zero. So from the ODZ we must exclude the values that turn the denominator to zero.
ODZ: x belongs to R\(0).
Now let's look at our equation as a proportion.
Basic 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;
Divide both sides of the equation by 100, thereby getting rid of the coefficient in front of the variable x:
Finding the unknown divisor
Let's look at the equation as a private one. Where the dividend is 5, the divisor is x, and the result of the division is the quotient is 100.
Recall the rule of how to find 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 of the equation. To do this, we substitute the found root into the original equation and perform the calculations:
The solution has been found correctly.
An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form
ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.
For example, all equations:
2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.
The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .
For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. Hence, the value x \u003d 2 is the solution or the root of the equation.
And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 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.
We transfer the free term from the left side of the equation to the right, while 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.
We transfer 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 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.
So the value x = 3 is the solution or the root of the equation.
Answer: x = 3.
If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 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 \u003d 3x - 12 + 2x - 1.
5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.
Here are similar members:
0x = 0.
Answer: x is any number.
If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.
Example 3 Solve the equation x + 8 = x + 5.
Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.
Here are similar members:
0x = - 3.
Answer: no solutions.
On figure 1 the scheme for solving the linear equation is shown
Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.
Example 4 Let's 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 members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar members:
- 22x = - 154.
6) Divide by - 22 , We get
x = 7.
As you can see, the root of the equation is seven.
In general, such equations can be solved as follows:
a) bring the equation to an integer form;
b) open 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 like terms.
However, this scheme is not required for every equation. When solving many simpler equations, one has to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5 Solve the equation 2x = 1/4.
We find the unknown x \u003d 1/4: 2,
x = 1/8 .
Consider the solution of some linear equations encountered in the main state exam.
Example 6 Solve 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) \u003d 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 \u003d 6 - 2, x \u003d 4.
If x = 4 then
f(6) = 3 7-4 = 3 3 = 27
Answer: 27.
If you still have questions, there is a desire to deal with the solution of 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 tutorial from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.
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