Number of solutions for system of equations

[Guest]

Hello!

I have a simple question about solutions, better said number of solutions for this system of equations.

[tex]\begin{cases}
x_{1 } − x_{2 } + 3x_{3 } − 2x_{4 } = 1\\
−2x_{1 } + 2cx_{2 } − 4x_{3 } + 2x_{4 } = −7\\
− 2x_{3 } + (−c + 6)x_{4 } = 2c + 15\\
− 2x_{3 } + c^{2 }x_{4 } = c^{2 }\end{cases}[/tex]

I know it's only possible that this system has either 0, 1 or [tex]\infty[/tex] number of solutions, for different values of c:
[tex]c = -3 \rightarrow \infty\\
c = 1 \rightarrow \infty\\
c = 2 \rightarrow 0 \\
c \in ℝ \setminus \{-3, 1, 2\} \rightarrow 1[/tex]

My question is: for which c has this system at the utmost 2 solutions? Should it be only for when the whole system has only one solution or also when it has none? Thank you for your help!

[Guest]

You say you know that such a system of equations must have no solution, one solution or infinitely many solutions so "at the utmost two solutions". "At the utmost" means "not more than" so here it must mean 0 or one solution and you have already answered that- this system (according to you, I have not checked myself) has 0 solutions for c= 2 and 1 solution for c any real number except -3, 1, or 2. So this system of equations has "at utmost two solutions" for c any real number except -3 or 1.

[Guest]

aI would do this problem by setting it up its coefficient matrix
[tex]\begin{bmatrix} 1 & -1 & 3 & -1 & 1 \\ 2 & 2c & 4 & 2 & -7 \\ 0 & 0 & -2 & 6- c & 2c+ 15 \\ 0 & 0 & -2 & c^2 & c^2\end{bmatrix}[/tex]
and row-reduce to a diagonal matrix by subtracting twice the first row from the second row and subtracting the third row from the fourth row:
[tex]\begin{bmatrix} 1 & -1 & 3 & -2 & 1 \\0 & 2c+ 2 & -2 & 6 & -9 \\ 0 & 0 & 0 & c^2+ c- 6 & c^2- 2c- 15\end{bmatrix}[/tex]

Now the question of how many solutions there are depends entirely on that last row. If [tex]c^2- 2c- 15[/tex] is non-zero then [tex]x_4= \frac{c^2- 2c- 15}{c^2+ c- 6}[/tex] and the other x values follow from that. There is, in this case, a single solution.
If [tex]c^2+ c- 6= 0[/tex] then we cannot divide by it. In this case there are two possibilities. The equation 0x= a, for a non-zero, has NO solution because any x times 0 is 0. The equation 0x= 0 has infinitely many solutions for the same reason.
[tex]c^2+ c- 6= (c- 2)(c+ 3)= 0[/tex] has solutions c= 2 and c= -3. [tex]c^2- 2c- 15= (c- 5)(c+ 3)[/tex] has solutions c= 5 and c= -3.

Putting those together, if c is any number other than 2 or -3 this system of equations have exactly one solution. If c= 2, that last equation is [tex]0x_4= -15[/tex] which has no solution. This system of equations has no solution. If c= -3, that last equation is 0= 0 which is true for all [tex]x_4[/tex]. This system of equations has infinitely many solutions

[Guest]

The copper, zinc and nickel contents of the three different brasses
(A, B, C) are as follows:
A : (55 % Cu, 11 % Zn, 34 % Ni)
B : (48% Cu, 23 % Zn, 29 % Ni)
C : (51 % Cu, 10 % Zn, 39 % Ni)
How much of these alloys must be melted together to obtain about 14.0 kg of brass with 49.1 % copper, 19.4 % zinc and 31.5 % nickel?
[Cu, Zn, Ni] = ??

[Guest]

You want to find three quantities: the amount of metal A, which I will call "a", the amount of metal B, "b", the amount of metal C, "c", each in kilograms, so you need three equations.

Since A is 55% copper, B is 48% copper, C is 51% copper, and you want the result to be 49.1% copper, you must have 0.44a+ 0.48b+ 0.51c= 0.491(15)= 7.365.
(The "15" is the 15 kg total metal.)

Since A is 11% zinc, B is 23% zinc, C is 10% zinc, and you want the result to be 19.4% zinc, you must have 0.11a+ 0.23B+ 0.10C= 0.194(15)= 2.91.

Since A is 34% nickel, B is 29% nickel, C is 39% nickel, and you want the result to be 31.5% nickel, you must have 0.34a+ 0.29b+ 0.39c= 0.315(15)= 2.25.

[Guest]

I accidently thought the amount of metal was 15 kg but see that it was 14 kg so I am correcting that.

[quow]You want to find three quantities: the amount of metal A, which I will call "a", the amount of metal B, "b", the amount of metal C, "c", each in kilograms, so you need three equations.

Since A is 55% copper, B is 48% copper, C is 51% copper, and you want the result to be 49.1% copper, you must have 0.44a+ 0.48b+ 0.51c= 0.491(14)= 6.874.
(The "14" is the 14 kg total metal.)

Since A is 11% zinc, B is 23% zinc, C is 10% zinc, and you want the result to be 19.4% zinc, you must have 0.11a+ 0.23B+ 0.10C= 0.194(14)= 2.716.

Since A is 34% nickel, B is 29% nickel, C is 39% nickel, and you want the result to be 31.5% nickel, you must have 0.34a+ 0.29b+ 0.39c= 0.315(14)= 2.1.[/quote]

Also, I see that you end with
"[Cu, Zn, Ni] = ??"

This problem does NOT ask the amount of copper, zinc, and nickel. It asks for the amount of the three metals whose percentages of copper, zinc, and nickel are given.