Quadratic Simultaneous Equations with Four Variables
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I have the following equations, where $a$ to $j$ are real constants, and $w, x, y$ and $z$ are values to be solved for:
$$awx - bwy - cxy + dy^2 = 0$$
$$gyz - bwy - fwz + ew^2= 0$$
$$gyz - ixz - cxy + hx^2= 0$$
$$awx - ixz - fwz + jz^2 = 0$$
I have tried some elimination and substitution but that didn't seem to lead anywhere. Where should I go from here?
polynomials systems-of-equations quadratics
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
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add a comment |
$begingroup$
I have the following equations, where $a$ to $j$ are real constants, and $w, x, y$ and $z$ are values to be solved for:
$$awx - bwy - cxy + dy^2 = 0$$
$$gyz - bwy - fwz + ew^2= 0$$
$$gyz - ixz - cxy + hx^2= 0$$
$$awx - ixz - fwz + jz^2 = 0$$
I have tried some elimination and substitution but that didn't seem to lead anywhere. Where should I go from here?
polynomials systems-of-equations quadratics
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
$endgroup$
$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30
add a comment |
$begingroup$
I have the following equations, where $a$ to $j$ are real constants, and $w, x, y$ and $z$ are values to be solved for:
$$awx - bwy - cxy + dy^2 = 0$$
$$gyz - bwy - fwz + ew^2= 0$$
$$gyz - ixz - cxy + hx^2= 0$$
$$awx - ixz - fwz + jz^2 = 0$$
I have tried some elimination and substitution but that didn't seem to lead anywhere. Where should I go from here?
polynomials systems-of-equations quadratics
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
$endgroup$
I have the following equations, where $a$ to $j$ are real constants, and $w, x, y$ and $z$ are values to be solved for:
$$awx - bwy - cxy + dy^2 = 0$$
$$gyz - bwy - fwz + ew^2= 0$$
$$gyz - ixz - cxy + hx^2= 0$$
$$awx - ixz - fwz + jz^2 = 0$$
I have tried some elimination and substitution but that didn't seem to lead anywhere. Where should I go from here?
polynomials systems-of-equations quadratics
polynomials systems-of-equations quadratics
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
asked Dec 29 '18 at 1:12
Adam CavenderAdam Cavender
62
62
$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30
add a comment |
$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30
$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30
$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30
add a comment |
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$begingroup$
The only solution that i have found is $$w=x=y=z=0$$
$endgroup$
– Dr. Sonnhard Graubner
Dec 29 '18 at 8:30