0

I have the following 5 equations in 5 variables(A, B, C, D, E):

  1. $2 \times E \times A + D = 1 $
  2. $2\times E\times B + D = 0 $
  3. $D + E = 2$
  4. $A + B + C = 1$
  5. $A^2 + B^2 + C = \frac 7 4$

I know the solution is: $ (0 , \frac {-1} 2 , \frac 3 2 , 1, 1) $. Just don't know how to get there.

Adnan
  • 245
  • The system represents the intersection of three hyperquadrics and two planes. You can expect eight solutions, I guess, unless there is some kind of degeneracy. –  Jan 30 '21 at 17:20
  • 1
    there is another real solution $$\left(1, \frac{3}{2}, -\frac{3}{2}, 3, -1\right)$$ – Raffaele Jan 30 '21 at 17:22

2 Answers2

0

From the first two, $E\ne0$. After elimination* of $A, B, C, D$, you get

$$ \frac{E-1}{2E} \left( \frac{E-1}{2E} -1\right)+\frac{E-2}{2E} \left(\frac{E-2}{2E} -1\right)=\frac34.$$

or simply

$$E^2=1.$$

The rest follows.

*Use $D=2-E$, express $A, B$ in terms of $E$ and plug them in the difference of the last two.

0

\begin{align} 2e a+d&=1 \tag{1}\label{1} ,\\ 2 e b+d&=0 \tag{2}\label{2} ,\\ d+e&=2 \tag{3}\label{3} ,\\ a+b+c&=1 \tag{4}\label{4} ,\\ a^2+b^2+c&=\tfrac74 \tag{5}\label{5} . \end{align}

Substitution of $e=2-d$ from \eqref{3} into \eqref{1} and \eqref{2} leads to expression of $a$ and $b$ in terms of $d$:

\begin{align} a &= \tfrac12\frac{d-1}{d-2} \tag{6}\label{6} ,\\ b &= \tfrac12\frac{d}{d-2} \tag{7}\label{7} . \end{align}

Then, substitution of these expressions for $a,b$ into \eqref{4} gives the expression of $c$ in terms of $d$:

\begin{align} c&=-\tfrac32\frac1{d-2} \tag{8}\label{8} . \end{align}

substitution expressions for $a,b,c$ in terms of $d$ into \eqref{5} after simple rearrangements, results in equation is $d$:

\begin{align} -\tfrac54\frac{(d-1)(d-3)}{(d-2)^2} &=0 , \end{align}

so there are two solutions for $d$, $1$ and $3$, consequently, there are two solutions of \eqref{1}-\eqref{5}:

\begin{align} a &= 0 ,\quad b = -\tfrac12 ,\quad c = \tfrac32 ,\quad d=1 ,\quad e=1 ,\\ \text{and}\quad a &= 1 ,\quad b = \tfrac32 ,\quad c = -\tfrac32 ,\quad d=3 ,\quad e = -1 . \end{align}

g.kov
  • 13,581