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. Solving ninth grade equations involves the use of many different solving methods: graphical, algebraic addition methods, introducing new variables, using functions and converting equations from one type to a simpler one, and much more. The method for solving the equation is selected based on the initial data, so it is best to understand the methods clearly using examples.
Suppose we are given an equation of the following form:
\[\frac (18)(x^2-6x)-\frac(12)(x^2+6x)=\frac (1)(x)\]
To solve this equation, divide the left and right sides by \
\[\frac(18)(x-6)-\frac(12)(x+6)=1\]
\[\frac (6x+180)(x^2-36)=1\]
The resulting two roots are the solution to this equation.
Let's solve the equation:
\[ (x^2-2x)^2-3x^2+6x-4=0 \]
It is necessary to find the sum of all roots of this equation. To do this you need to replace:
The roots of this equation will be 2 numbers: -1 and 4. Therefore:
\[\begin(bmatrix) x^2-2x=-1\\ x^2-2x=4 \end(bmatrix)\] \[\begin(bmatrix) x=1\\ x=1\pm\sqrt5 \end(bmatrix)\]
The sum of all 3 roots is equal to 4, which will be the answer to solving this equation.
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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.
Solving 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:
- 22х = - 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 simpler equations, you have 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.
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.
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