Solving linear equations with examples
The algorithm for finding these points has already been discussed several times, but I’ll repeat it briefly:
1. Find the derivative of the function.
2. Find the zeros of the derivative (equate the derivative to zero and solve the equation).
3. Next, we build a number line, mark the found points on it and determine the signs of the derivative on the resulting intervals. *This is done by substituting arbitrary values from the intervals into the derivative.
If you are not at all familiar with the properties of derivatives for studying functions, then be sure to study the article« ». Also repeat the table of derivatives and the rules of differentiation (available in the same article). Let's consider the tasks:
77431. Find the maximum point of the function y = x 3 –5x 2 +7x–5.
Let's find the derivative of the function:
Let's find the zeros of the derivative:
3x 2 – 10x + 7 = 0
y(0)" = 3∙0 2 – 10∙0 + 7 = 7 > 0
y(2)" = 3∙2 2 – 10∙2 + 7 = – 1< 0
y(3)" = 3∙3 2 – 10∙3 + 7 = 4 > 0
At the point x = 1, the derivative changes its sign from positive to negative, which means this is the desired maximum point.
Answer: 1
77432. Find the minimum point of the function y = x 3 +5x 2 +7x–5.
Let's find the derivative of the function:
Let's find the zeros of the derivative:
3x 2 + 10x + 7 = 0
Deciding quadratic equation we get:
We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:
y(–3 ) " = 3∙(–3) 2 + 10∙(–3) + 7 = 4 > 0
y(–2 ) "= 3∙(–2) 2 + 10∙(–2) + 7 = –1 < 0
y(0) "= 3∙0 2 – 10∙0 + 7 = 7 > 0
At the point x = –1, the derivative changes its sign from negative to positive, which means this is the desired minimum point.
Answer: –1
77435. Find the maximum point of the function y = 7 + 12x – x 3
Let's find the derivative of the function:
Let's find the zeros of the derivative:
12 – 3x 2 = 0
x 2 = 4
Solving the equation we get:
*These are points of possible maximum (minimum) of the function.
We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:
y(–3 ) "= 12 – 3∙(–3) 2 = –15 < 0
y(0) "= 12 – 3∙0 2 = 12 > 0
y( 3 ) "= 12 – 3∙3 2 = –15 < 0
At the point x = 2, the derivative changes its sign from positive to negative, which means this is the desired maximum point.
Answer: 2
*For the same function, the minimum point is the point x = – 2.
77439. Find the maximum point of the function y = 9x 2 – x 3.
Let's find the derivative of the function:
Let's find the zeros of the derivative:
18x –3x 2 = 0
3x(6 – x) = 0
Solving the equation we get:
We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:
y(–1 ) "= 18 (–1) –3 (–1) 2 = –21< 0
y(1) "= 18∙1 –3∙1 2 = 15 > 0
y(7) "= 18∙7 –3∙7 2 = –1< 0
At the point x = 6, the derivative changes its sign from positive to negative, which means this is the desired maximum point.
Answer: 6
*For the same function, the minimum point is the point x = 0.
77443. Find the maximum point of the function y = (x 3 /3)–9x–7.
Let's find the derivative of the function:
Let's find the zeros of the derivative:
x 2 – 9 = 0
x 2 = 9
Solving the equation we get:
We determine the signs of the derivative of the function on intervals and mark them on the sketch. We substitute an arbitrary value from each interval into the derivative expression:
y(–4 ) "= (–4) 2 – 9 > 0
y(0) "= 0 2 – 9 < 0
y(4) "= 4 2 – 9 > 0
At the point x = – 3, the derivative changes its sign from positive to negative, which means this is the desired maximum point.
Answer: – 3
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 compose general scheme 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.
website, when copying material in full or in part, a link to the source is required.
It consists in the fact that concrete, reinforced with strong steel frames, is highly durable building material and is not subject to numerous influences environment, due to which the design of the overhead line support foundation is capable of supporting steel and reinforced concrete power transmission line supports without the threat of them toppling over for decades. Durability, load resistance and strength are the main advantages of using products reinforced concrete foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines in energy construction.
Reinforced concrete foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines are made of heavy concrete with a compressive strength class not lower than B30, grade - from M300. The grade of concrete for frost resistance is not lower than F150, for water resistance - W4 - W6. Cement and inerts used for the production of concrete must meet the requirements of SNiP I-B.3-62 and TP4-68. Largest size grains in the concrete structure should not exceed 20-40 mm. Control of concrete strength of support foundations in accordance with GOST 10180-67 “Heavy concrete. Methods for determining strength" and GOST 10181-62 "Heavy concrete. Methods for determining the mobility and rigidity of a concrete mixture."
The following are used as reinforcement for foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines: hot-rolled reinforcing steel bars class A-I, bar hot-rolled reinforcing steel of periodic profile class A-III, bar reinforcing steel of periodic profile of class A-IV and ordinary reinforcing wire of class B1. For mounting loops, only hot-rolled rod reinforcement of class A-I made of carbon mild steel is used.
The foundations of power transmission line supports for energy construction face a responsible task - to maintain the stability and strength of power transmission line supports for many years in different climatic conditions, at any time of the year and in any weather. Therefore, very high demands are placed on the foundations of supports. high demands. Before shipping to the customer, the foundations of FP2.7x2.7-A5 supports for metal supports of 500 kV overhead lines are tested according to various parameters, for example, the degree of stability, strength, durability and wear resistance, resistance negative temperatures and atmospheric influences. Before welding, joint parts must be free of rust. Reinforced concrete foundations with a concrete protective layer thickness of less than 30 mm, as well as foundations installed in aggressive soils, must be protected with waterproofing.
During operation, foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines are subject to careful supervision, especially in the first years of operation of the overhead line. One of the most serious defects in the construction of foundations, difficult to eliminate under operating conditions, is a violation of technological standards during their manufacture: the use of low-quality or poorly washed gravel, violation of proportions when preparing a concrete mixture, etc. An equally serious defect is layer-by-layer concreting of foundations, when individual elements of the same foundation are concreted in different times without preliminary preparation surfaces. In this case, the concrete of one foundation element does not set to another and the foundation may be destroyed under external loads that are significantly less than the calculated ones.
When making reinforced concrete foundations for supports, standards are also sometimes violated: low-quality concrete is used, reinforcement is laid in the wrong sizes as provided for in the project. During the construction of power lines on prefabricated or piled reinforced concrete foundations, serious defects may occur that are not allowed by energy construction. Such defects include the installation of broken reinforced concrete foundations, their insufficient penetration into the ground (especially when installing supports on the slopes of hills and ravines), inappropriate compaction during backfilling, installation of prefabricated foundations of smaller sizes, etc. Installation defects include incorrect installation of reinforced concrete foundations, in which individual prefabricated foundations intended as the base of a metal support have different vertical elevations or shifts of individual foundations in plan. If unloaded incorrectly, the FP2.7x2.7-A5 foundations for metal supports of 500 kV overhead lines may be damaged, concrete may chip and reinforcement may be exposed. In the process of acceptance special attention should be ensured that the anchor bolts and their nuts correspond to the design dimensions.
Under operating conditions, reinforced concrete foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines are damaged both from impacts external environment, and from large external loads. The reinforcement of foundations with a porous concrete structure is damaged by the aggressive effects of groundwater. Cracks that form on the surface of foundations, when exposed to operational alternating loads, as well as wind, moisture and low temperature, expand, which ultimately leads to the destruction of concrete and exposure of reinforcement. In areas located near chemical plants, anchor bolts quickly deteriorate and upper part metal footrests.
Breakage of the foundation of supports can also occur as a result of its misalignment with the racks, which causes the appearance of large bending moments. A similar breakdown can occur when the base of the foundation is washed away by groundwater and deviates from its vertical position.
During the acceptance process, foundations FP2.7x2.7-A5 for metal supports of 500 kV overhead lines are checked for their compliance with the design, laying depth, quality of concrete, quality of welding of working reinforcement and anchor bolts, presence and quality of protection against the action of aggressive waters. The vertical marks of the foundations are measured and the location of the anchor bolts is checked according to the template. If any non-compliance with the standards is detected, all defects are eliminated before backfilling the pits. Foundations that have chipped concrete and exposed reinforcement in the upper part are repaired. To do this, a concrete frame 10-20 cm thick is installed, buried 20-30 cm below ground level. It should be borne in mind that energy construction does not allow a frame made of slag concrete, since the slag contains an admixture of sulfur, which causes intense corrosion of reinforcement and anchors. bolts In case of more significant damage to foundations (including monolithic ones), the damaged part is covered with reinforcement welded to the reinforcement of the main foundation, and after installation of the formwork it is concreted.