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I've stuck in this equation system. No clue how to start ?

$$\begin{eqnarray} x+y+z &=&a+b+c\tag{1} \\ ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\ ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3} \end{eqnarray}$$

Find the value of $x,y,z$ is in the form of $a,b$ and $c$. I want to know steps of solution.

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2  
Are $a,b,c$ given? What are the dots-just for spacing?, $x=a,y=b,z=c$ looks easy to find. One would expect five more solutions. –  Ross Millikan May 24 '13 at 18:30

3 Answers 3

I'm supposed to solve it using simultaneous eqn method like elimination,substitution,cross-multiplication .NO knowledge of matrices

By inspection we see that $(x,y,z)=(a,b,c)$ is a solution of the given system

$$ \begin{array}{l} \text{Eq. 1}\qquad x+y+z=a+b+c \\ \text{Eq. 2}\qquad ax+by+cz=a^{2}+b^{2}+c^{2} \\ \text{Eq. 3}\qquad ax^{2}+by^{2}+cz^{2}=a^{3}+b^{3}+c^{3}. \end{array} \tag{0}$$

The other solution can be found as follows. Solve Eq. 1 for $z$. Multiply original Eq. 1 by $a$, subtract Eq. 2 and solve for $z$. This results in $$ \begin{array}{l} z=a+b+c-x-y, \end{array} \tag{1}$$

$$ \begin{array}{l} z=\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c} . \end{array} \tag{2}$$

Equate the right hand sides of $(1)$ and $(2)$

$$ \begin{array}{l} \frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}=a+b+c-x-y, \end{array} \tag{3}$$

and solve for $x$

$$ x=\frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}. \tag{4}$$

Substitute $x,z$ in $(0)$, Eq. 3, using $(4)$ and $(2)$

$$ a\left( \frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}\right) ^{2}+by^{2}+c\left( \frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}\right) ^{2}=a^{3}+b^{3}+c^{3}. \tag{5}$$

Solving for $y$ we get$^1$ the solution $y=b$ and the solution

$$y=\frac{B}{D},\tag{6}$$

where

$$ B=-2a^{3}c+2a^{3}b-a^{2}b^{2}-a^{2}bc+4a^{2}c^{2}-2acb^{2}-2ac^{3}+ab^{3}-abc^{2}+2bc^{3}+b^{3}c-c^{2}b^{2}$$ $$ D=a^{2}c+ac^{2}-6abc+a^{2}b+bc^{2}+ab^{2}+b^{2}c $$

Finally substitute $y=b$ and $y=B/D$ in $(4)$ and $(2)$. We get the solutions $(x,z)=(a,c)$, and

$$(x,z)=\left(\frac{A}{D},\frac{C}{D}\right),\tag{7}$$

where

$$ A=a^{3}c+a^{3}b-2a^{2}bc-a^{2}b^{2}-a^{2}c^{2}+2ac^{3}-abc^{2}+2ab^{3}-acb^{2}-2bc^{3}-2b^{3}c+4c^{2}b^{2} $$

$$ C=2a^{3}c-2a^{3}b+4a^{2}b^{2}-a^{2}c^{2}-a^{2}bc-acb^{2}-2ab^{3}+ac^{3}-2abc^{2}+bc^{3}+2b^{3}c-c^{2}b^{2}. $$

So the two solutions of $(0)$ are: $$(x,y,z)=(a,b,c)\qquad\text{and }\qquad(x,y,z)=\left(\frac{A}{D},\frac{B}{D},\frac{C}{D}\right).$$

--

$^1$ Eq. $(5)$ is equivalent to

$$\begin{equation*} \left( cb^{2}+c^{2}b+ac^{2}+ca^{2}+ab^{2}-6acb+ba^{2}\right) (y-b)\left( y-\frac{B}{D}\right) =0. \end{equation*}$$

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As a side note, you might want to assume that $a,b,c$ are distinct, so that you can use them in your denominator. If $a=b$, then there is only 1 solution (and your other predicted solution simplifies to $x=a,y=b,z=c$.) –  Calvin Lin May 24 '13 at 22:24
    
@Calvin Lin Thanks. In the meantime I've edited my answer. –  Américo Tavares May 24 '13 at 22:30
    
@CalvinLin I like the geometric interpretation in your answer. –  Américo Tavares May 24 '13 at 22:40

Assume that $(a,b,c)\ne \lambda(1,1,1)$ and put $x:=a+u$, $\>y:=b+v$, $\>z:=c+w$. Then $(1)$ and $(2)$ amount to the system $$\eqalign{\phantom{a}u+\phantom{b}v+\phantom{c}w&=0\cr a u+ bv+cw&=0\cr}$$ with the solution $$u=t(b-c),\quad v=t(c-a),\quad w=t(b-c)\qquad(t\in{\mathbb R})\ .$$ Plug this into the equation $(3)$: $$a(a+u)^2+b(b+v)^2+c(c+w)^2=a^3+b^3+c^3$$ and obtain a quadratic equation for $t$ with one obvious solution $t=0$ and a second solution $$t={2(a-b)(b-c)(c-a)\over ab(a+b)+bc(b+c)+ca(c+a)-6abc}\ .\tag{4}$$ Cases where the denominator in $(4)$ is zero have to be dealt with separately.

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This is a slightly interesting question. Let me try and provide some motivation for the ugly values that Americo calculated.

Let me assume that $a\neq b, b \neq c, c \neq a $. We'd deal with this issue separately.

Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which we can calculate. The first 2 equations are equivalent to $$ \begin{array} {llll} &&(x-a) &+ &(y-b) &+ &(z-c) &= 0\\ &a&(x-a) &+b&(y-b) & +c&(z-c) & = 0 \\ \end{array}$$

Since

$$ \begin{pmatrix} 1\\1\\1\\ \end{pmatrix} \times \begin{pmatrix} a\\b\\c\\ \end{pmatrix} = \begin{pmatrix} c-b\\a-c\\b-a\\ \end{pmatrix}, $$

we know that the line is given by $\frac{x-a}{c-b} = \frac{ y-b}{a-c} = \frac{z-c}{b-a} = k $. (This is valid because the denominator is never 0.) This is equivalent to $x=a + k(c-b), y = b+k(a-c) , z = c+k(b-a)$.

Next, we are interested in when this line intersects the ellipsoid $ax^2 + by^2 + cz^2 = a^3 + b^3 + z^3 $, which happens in at most 2 places. (Of course, one place is $k=0$, which leads to the solution $x=a, y=b, z=c$.) Using the equation of the line above, we get that

$$\begin{array}{l l l l} ax^2 &= a^3 & + 2a^2k(c-b) &+k^2 a(c-b)^2 \\ by^2 &= b^3 & + 2b^2k(a-c) & + k^2 b(a-c)^2 \\ cz^2 &= c^3 & + 2c^2k(b-a) & + k^2 c(b-a)^2 \\ \end{array} $$

Summing up these three equations, and using the last equation given, we get that

$$-2k(a-b)(b-c)(c-a) + k^2 (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc) = 0. $$

One solution is $k=0$ as mentioned, and the other is

$$ k = \frac{ 2 (a-b)(b-c)(c-a) } { a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc} .$$

With this value of $k$, we get the answer that Americo calculated above.


Now, what happens when $a=b$ or $b=c$ or $c=a$? Interestingly, this forces there to be only 1 solution, namely $k=0$. You can also get this result by substituting $a=b$ into the value of $k$ and showing that the numerator is 0. The geometric interpretation is that the line becomes tangential to the ellipsoid.

Similarly, if $a=b=c$, then there is only 1 solution.

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