glitinosim3

Answered

2022-06-30

mixed system of equalities and inequalities, for example:

$x+2y+3z=10\phantom{\rule{0ex}{0ex}}2x+4y+10z=20\phantom{\rule{0ex}{0ex}}4x+y+z<10$

where unknown variables $x,y,z$ are all real-valued.

How can get a solution (or the range of all feasible solutions) for this system?

$x+2y+3z=10\phantom{\rule{0ex}{0ex}}2x+4y+10z=20\phantom{\rule{0ex}{0ex}}4x+y+z<10$

where unknown variables $x,y,z$ are all real-valued.

How can get a solution (or the range of all feasible solutions) for this system?

Answer & Explanation

Jordin Church

Expert

2022-07-01Added 11 answers

Multiply the first equation with 2, the eq.1 and eq. 2 read:

$2x+4y+6z=20\phantom{\rule{0ex}{0ex}}2x+4y+10z=20\phantom{\rule{0ex}{0ex}}$

Subtraction gives: $4z=0$ hence $z=0.$

From the first eq. we now derive $x=10-2y.$ Thus the Third eq. becomes: $4(10-2y)+y<1=$ or $y>\frac{30}{7}.$

Consequence: the set of solutions is given by:

$\{(10-2y,y,0):y>\frac{30}{7}\}.$

$2x+4y+6z=20\phantom{\rule{0ex}{0ex}}2x+4y+10z=20\phantom{\rule{0ex}{0ex}}$

Subtraction gives: $4z=0$ hence $z=0.$

From the first eq. we now derive $x=10-2y.$ Thus the Third eq. becomes: $4(10-2y)+y<1=$ or $y>\frac{30}{7}.$

Consequence: the set of solutions is given by:

$\{(10-2y,y,0):y>\frac{30}{7}\}.$

Lucian Maddox

Expert

2022-07-02Added 8 answers

I see, thank you. My problem is my system is usually composed of M equations and N inequalities (and M+N variables) where both M and N can be very large, making an analytical solution potentially tedious

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