Instructions
Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without the “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.
Do a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!
A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in the equation “y” has a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. fold the left side with the left, and the right with the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2 Using the 1st method, you can check that the roots are found correctly.
If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have “2x”, and in the second we simply have “x”. To reduce x when adding or subtracting, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus in front of the bracket, then after opening, change the signs to the opposite ones:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4
Video on the topic
When solving differential equations, the argument x (or time t in physical problems) is not always explicitly available. However, this is a simplified special case specifying a differential equation, which often helps to simplify the search for its integral.
Instructions
Consider a physical problem that results in differential equation, in which the argument t is missing. This is a problem about oscillations of a mass m suspended on a thread of length r located in a vertical plane. The equation of motion of the pendulum is required if it was initially motionless and tilted from the equilibrium state by an angle α. The forces should be neglected (see Fig. 1a).
Solution. A mathematical pendulum is a material point suspended on a weightless and inextensible thread at point O. Two forces act on the point: the force of gravity G=mg and the tension force of the thread N. Both of these forces lie in the vertical plane. Therefore, to solve the problem we can apply the equation rotational movement points around a horizontal axis passing through point O. The equation of rotational motion of a body has the form shown in Fig. 1b. In this case, I is the moment of inertia of the material point; j is the angle of rotation of the thread together with the point, measured from the vertical axis counterclockwise; M is the moment of forces applied to a material point.
Calculate these values. I=mr^2, M=M(G)+M(N). But M(N)=0, since the line of action of the force passes through point O. M(G)=-mgrsinj. The “-” sign means that the moment of force is directed in the direction opposite to the movement. Substitute the moment of inertia and the moment of force into the equation of motion and get the equation shown in Fig. 1s. By reducing the mass, a relationship emerges (see Fig. 1d). There is no t argument here.
Textbook:
- Makarychev Yu. N., Mindyuk N. R. Mathematics. 7th grade
Goals:
I. Student survey
- What is a function called?
- What is the domain of a function?
- What is the range of a function?
- What functions did we get to know?
- What is the graph of a linear function? ( straight). How many points are needed to construct this graph?
(A function is the dependence of one variable on another, in which each value of the independent variable corresponds to a single value of the dependent variable)
(All values that the independent variable (argument) takes form the domain of the function.)
(All values that the dependent variable takes are called function values)
a) with a linear function of the form y = kx + b,
direct proportionality of the form y = kx
b) with functions of the form y = x 2, y = x 3
Without performing construction, determine the relative position of the graphs of functions given by the following formulas:
A ) y = 3x + 2; y = 1.2x + 5;
b) y = 1.5x + 4; y = -0.2x + 4; y = x + 4;
With) y = 2x + 5; y = 2x - 7; y = 2x
Picture 1
The figure shows graphs of linear functions ( Each student is given a sheet of paper with the graphs at their desk.). Write a formula for each graph
What function graphs are we still familiar with? ( y = x 2; y = x 3 )
- What is the graph of a function y = x 2 (parabola).
- How many points do we need to construct to depict a parabola? ( 7, one of which is the vertex of a parabola).
Let's build a parabola given by the formula y = x 2
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y = x 2 | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
| y = x 2 + 2 | 11 | 6 | 3 | 2 | 3 | 6 | 11 |
Figure 2
What properties does the graph of a function have? y = x 3 ?
- If x = 0 , That y = 0 - vertex of the parabola (0;0)
- Domain: X - any number, D (y) = (- ?; ?) D (y) = R
- Range of values at ? 0
- E (y) =
- The function increases over the interval
The function increases over the interval - at these values of x, moving along the parabola from left to right, we “go down the hill” (see Fig. 55). The function y = x 2 increases along the ray;
b) on the segment [- 3, - 1.5];
c) on the segment [- 3, 2].Solution,
a) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values of the variable x from the segment (Fig. 56). For the selected part of the graph we find at the name. = 1 (at x = 1), y max. = 9 (at x = 3).
b) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values of the variable x from the segment [-3, -1.5] (Fig. 57). For the selected part of the graph, we find y name. = 2.25 (at x = - 1.5), y max. = 9 (at x = - 3).
c) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values of the variable x from the segment [-3, 2] (Fig. 58). For the selected part of the graph, we find y max = 0 (at x = 0), y max. = 9 (at x = - 3).
Advice. To avoid plotting the function y - x 2 point by point each time, cut out a parabola template from thick paper. With its help you will draw a parabola very quickly.
Comment. By inviting you to prepare a parabola template, we seem to be equalizing the rights of the function y = x 2 and linear function y = kx + m. After all, the schedule linear function is a straight line, and to depict a straight line, an ordinary ruler is used - this is the template for the graph of the function y = kx + m. So let you have a template for the graph of the function y = x 2.
Example 2. Find the intersection points of the parabola y = x 2 and the straight line y - x + 2.
Solution. Let us construct in one coordinate system the parabola y = x 2 and the straight line y = x + 2 (Fig. 59). They intersect at points A and B, and from the drawing it is not difficult to find the coordinates of these points A and B: for point A we have: x = - 1, y = 1, and for point B we have: x - 2, y = 4.
Answer: the parabola y = x 2 and the straight line y = x + 2 intersect at two points: A (-1; 1) and B (2; 4).
Important note. Until now, we have been quite bold in drawing conclusions using the drawing. However, mathematicians do not trust drawings too much. Having discovered in Figure 59 two points of intersection of a parabola and a straight line and having determined the coordinates of these points using the drawing, the mathematician usually checks himself: whether the point (-1; 1) actually lies on both the straight line and the parabola; does the point (2; 4) really lie on both a straight line and a parabola?
To do this, you need to substitute the coordinates of points A and B into the equation of the straight line and into the equation of the parabola, and then make sure that in both cases the correct equality is obtained. In example 2 in both cases we get true equalities. This check is especially often carried out when there is doubt about the accuracy of the drawing.
In conclusion, we note one interesting property of the parabola, discovered and proven jointly by physicists and mathematicians.
If we consider the parabola y = x 2 as a screen, as a reflective surface, and place a light source at the point, then the rays, reflected from the parabola of the screen, form parallel beam light (Fig. 60). The point is called the focus of the parabola. This idea is used in cars: the reflective surface of the headlight has a parabolic shape, and the light bulb is placed at the focal point - then the light from the headlight spreads far enough.
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“Graph of function Y X”- From the above it follows that the graph of the function y=(x - m)2 + n is a parabola with the vertex at the point (m; n). Construct the function graphs yourself: y = x2 + 2; y = x2 – 3; y = (x – 1)2; y = (x + 2)2; y = (x + 1)2 – 2; y = (x – 2)2 + 1; y = (x + 3)*(x – 3); y = x2 + 4x – 4; y = x2 – 6x + 11. The graph of the function y=(x - m)2 is a parabola with its vertex at point (m; 0).