![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
3x+2y+z=14 { x+y+z=10 2x+3y-z=1
① 3x + 2Y + Z = x + y + Z + 2x + y = 10 + 2x + y = 14, that is, 2x + y = 4, y = 4-2x
② From 2x + 3y-z = 1: 3Y = 1 + z-2x = 3 (4-2x). After finishing: z = 11-4x
③ Substituting y and Z in 1 and 2 into the original formula, we get x = 1, y = 2, z = 7
④ Substituting into the original formula: 3 × 1 + 2 × 2 + 7 = 14, the calculation is correct
3x+2y=14,x+y+z=10,2x+3y-z=1
Known:
3x+2y=14 ①
x+y+z=10 ②
2x+3y-z=1 ③
②+③=x+y+2x+3y=10+1
3x+4y=11 ④
④-①=（3x+4y）-（3x+2y）=11-14
2y= - 3
y= -1.5
Substitute: x = 17 / 3
z=35/6
3x+2y+z=14 x+y+2z+-3 2x+3y-z=1
(1) Ten (3) cancels Z to get x ten y = 3
Substitute (2) to get the value of Z, and (2) has a problem
RELATED INFORMATIONS
- 1. x+y+z=4 2x+3y-z=6 3x+2y+2z=10
- 2. The second power of [(x + 3Y) (x-3y)]
- 3. How to solve the equation axdy = xdx + YDY? How to solve the differential equation axdy = xdx + YDY? You don't need to calculate the specific value, just find out the relationship between X and y Give a specific process, thank you
- 4. The U / V power of Z = UE u = x ^ 2 + y ^ 2 v = XY
- 5. It is proved that the differentiable function must be continuous
- 6. Seeking total differential
- 7. Finding the range of function y = x2-2x + 2
- 8. Quadratic function problems in the third grade of Mathematics In the plane rectangular coordinate system, OA = 12 cm and ob = 6 cm are known. Point P starts from point O and moves along the edge of OA to point a It moves at a speed of 1 cm / S; point Q moves from B along Bo to o at a speed of 1 cm / s P. Q starts at the same time, and t (second) represents the moving time (0 ≤ T)
- 9. Analytic expression of quadratic function I'm a bit at a loss when I learn this quadratic function. I don't know how to start. Please give me some examples of different methods, that is, different conditions, and then write the answers to show me. Thank you very much,
- 10. It is proved that the second order linear ordinary differential equation has two linearly independent solutions The equation is as follows: y '' + P (x) * y '+ Q (x) * y = 0; It is proved that the differential equation must have two linearly independent solutions; How to prove it? Why must it be two? And linearly independent?
- 11. Evaluation: x + 3Y divided into x-2y, 3x-5y = 0
- 12. A calculus problem f (x + y, X-Y) = XY, then what are the partial derivatives of F for X and F for y?
- 13. Finding the total differential of Z = in (XY) + e ^ (x + y ^ 2)
- 14. Finding the total differential of Z = XY + (x / y) at point (1,2)
- 15. Given function FX satisfies f (x + y) = f (x) + F (y) for any XY ∈ R Find the value of 1 f (0) 2 F (x) is an odd function
- 16. Find the derivative and differential of function y = x ^ 3tan ^ 2 (x ^ 2-4)
- 17. The concept of function is based on mapping, but mapping does not allow one to many. Why does the so-called multi valued function allow one to many? Based on your answers, I come to the conclusion that: Generally, function is the abbreviation of single valued function, which is built on the basis of mapping. Multi valued function is a different concept system from function, that is, function is not the general name of single valued function and multi valued function. Multi valued function is an independent concept, which is not built on the basis of mapping. Its definition stipulates that it can be one to many I don't know if I can say that
- 18. Several concepts of advanced numbers, If the function f (x, y) has a minimum at point (x0, Y0), then (x0, Y0) must be a function of F (x, y) The minimum value of a continuous point B in the domain of definition C the minimum value of stationary point D in a certain field (x0, Y0) Let f (x), G (x) be continuous on [a, b], and f (x) ≥ g (x), then A ∫ (upper limit B, lower limit a) f (x) DX ≥ ∫ (B, a) g (x) DX B ∫(b,a)f(x)dx≤∫ (b,a)g(x)dx C ∫ f(x)dx≥∫ g(x)dx D ∫ f(x)dx=∫ g(x)dx What is the divergence of the following generalized integrals A ∫ (upper limit + ∞, lower limit 0) DX / 1 + x ^ 2 B ∫ (1,0) DX / radical 1-x ^ 2 C ∫(+∞,e)(lnx/x)dx D ∫(+∞,e)e^-x dx The continuity of function f (x, y) in P 0 (x 0, y 0) is the existence of the first partial derivative of F (x, y) in P 0 (x 0, y 0) A necessary condition B sufficient condition C necessary and sufficient condition D neither necessary nor sufficient condition Let f (x, y) = {(x ^ 2) y / x ^ 4 + y ^ 2 (x, y) not = (0,0); 0 (x, y) = (0,0) Alim (x, y) → (0,0) f (x, y) exists and f (x, y) is discontinuous at (0,0) B LIM (x, y) → (0,0) f (x, y) does not exist, f (x, y) is discontinuous at (0,0) C LIM (x, y) → (0,0) f (x, y) exists and f (x, y) is continuous at (0,0) D LIM (x, y) → (0,0) f (x, y) does not exist, f (x, y) is continuous at (0,0) Let a = ∫ (2,1) lnxdx, B = ∫ (2,1) | LNX | DX, then A a=b B a＞b C a＜b D a≥h
- 19. The meaning of differential, Derivative is the rate of change, integral is the magnitude of change, what about differential? In addition, what is the significance of studying differential?
- 20. X = u + V y = u ^ 2 + V ^ 2 z = u ^ 3 + V ^ 3 finding the partial differential of Z to X