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kipiarov [429]
3 years ago
11

A system of three linear equations in three variables is inconsistent. how many solutions to the system exist? none one three in

finitely many
Mathematics
1 answer:
Illusion [34]3 years ago
3 0

No solutions exist to this system of equations

A system of equations is said to be inconsistent if the two equations do not agree, meaning that there is no set of values that can simultaneously satisfy each equation. For example, the most basic inconsistent system of equations could be x = 4 a nd x = 5 at same time which is not possible.

A single equation in three unknowns can be interpreted geometrically in 3-dimensional space. If we look at a system of such linear equations, the solution set is the set of all points lying in all of the corresponding planes.

Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location.

Hence, No solutions exist to this system of equations

Learn more about inconsistent equations here :

brainly.com/question/5305250

#SPJ4

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The vertices of polygon ABCD are at A(1, 1), B(2, 3), C(3, 2), and D(2, 1). ABCD is reflected across the x-axis and translated 2
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Read 2 more answers
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

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