Mixing
How to solve the following system of equations?
$begingroup$
$left{ begin{aligned} xy + 2x + 2y &= -8\ yz + 2y + 2z &= 24\ xz + 2x + 2z &= -11 end{aligned} right.$
I need to solve it over the set of real numbers.
systems-of-equations quadratics symmetric-polynomials
edited Jan 2 at 12:49
Harry Peter
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
$endgroup$
add a comment |
$begingroup$
$left{ begin{aligned} xy + 2x + 2y &= -8\ yz + 2y + 2z &= 24\ xz + 2x + 2z &= -11 end{aligned} right.$
I need to solve it over the set of real numbers.
systems-of-equations quadratics symmetric-polynomials
edited Jan 2 at 12:49
Harry Peter
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
$endgroup$
add a comment |
$begingroup$
$left{ begin{aligned} xy + 2x + 2y &= -8\ yz + 2y + 2z &= 24\ xz + 2x + 2z &= -11 end{aligned} right.$
I need to solve it over the set of real numbers.
systems-of-equations quadratics symmetric-polynomials
edited Jan 2 at 12:49
Harry Peter
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
$endgroup$
$left{ begin{aligned} xy + 2x + 2y &= -8\ yz + 2y + 2z &= 24\ xz + 2x + 2z &= -11 end{aligned} right.$
I need to solve it over the set of real numbers.
systems-of-equations quadratics symmetric-polynomials
systems-of-equations quadratics symmetric-polynomials
edited Jan 2 at 12:49
Harry Peter
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
edited Jan 2 at 12:49
Harry Peter
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
edited Jan 2 at 12:49
Harry Peter
5,47111439
edited Jan 2 at 12:49
Harry Peter
5,47111439
edited Jan 2 at 12:49
Harry Peter
5,47111439
5,47111439
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
asked Dec 30 '18 at 19:25
Krisztián KissKrisztián Kiss
313
313
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Note that $$(x+2)(y+2)=xy+2x+2y+4$$Hence, our system of equations becomes $$(x+2)(y+2)=-4$$$$(y+2)(z+2)=28$$$$(x+2)(z+2)=-7$$Can you solve from here?
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
$endgroup$
add a comment |
$begingroup$
Elimimating the variable $x$ from (1) and plug this in (3) we get
$$-frac{2 (y+4) z}{y+2}-frac{4 (y+4)}{y+2}+2 z+11=0$$
Now we get $z$ from (2) $$z=-frac{2 (y-12)}{y+2}$$ finally we obtain one equation in $y$ factorizing we get $$frac{7 (y-2) (y+6)}{(y+2)^2}=0$$ From here we find the Solutions of our System.
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
Note that $$(x+2)(y+2)=xy+2x+2y+4$$Hence, our system of equations becomes $$(x+2)(y+2)=-4$$$$(y+2)(z+2)=28$$$$(x+2)(z+2)=-7$$Can you solve from here?
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
$endgroup$
add a comment |
$begingroup$
Note that $$(x+2)(y+2)=xy+2x+2y+4$$Hence, our system of equations becomes $$(x+2)(y+2)=-4$$$$(y+2)(z+2)=28$$$$(x+2)(z+2)=-7$$Can you solve from here?
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
$endgroup$
add a comment |
$begingroup$
Note that $$(x+2)(y+2)=xy+2x+2y+4$$Hence, our system of equations becomes $$(x+2)(y+2)=-4$$$$(y+2)(z+2)=28$$$$(x+2)(z+2)=-7$$Can you solve from here?
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
$endgroup$
Note that $$(x+2)(y+2)=xy+2x+2y+4$$Hence, our system of equations becomes $$(x+2)(y+2)=-4$$$$(y+2)(z+2)=28$$$$(x+2)(z+2)=-7$$Can you solve from here?
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
answered Dec 30 '18 at 19:29
Don ThousandDon Thousand
4,271734
4,271734
add a comment |
add a comment |
$begingroup$
Elimimating the variable $x$ from (1) and plug this in (3) we get
$$-frac{2 (y+4) z}{y+2}-frac{4 (y+4)}{y+2}+2 z+11=0$$
Now we get $z$ from (2) $$z=-frac{2 (y-12)}{y+2}$$ finally we obtain one equation in $y$ factorizing we get $$frac{7 (y-2) (y+6)}{(y+2)^2}=0$$ From here we find the Solutions of our System.
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Elimimating the variable $x$ from (1) and plug this in (3) we get
$$-frac{2 (y+4) z}{y+2}-frac{4 (y+4)}{y+2}+2 z+11=0$$
Now we get $z$ from (2) $$z=-frac{2 (y-12)}{y+2}$$ finally we obtain one equation in $y$ factorizing we get $$frac{7 (y-2) (y+6)}{(y+2)^2}=0$$ From here we find the Solutions of our System.
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Elimimating the variable $x$ from (1) and plug this in (3) we get
$$-frac{2 (y+4) z}{y+2}-frac{4 (y+4)}{y+2}+2 z+11=0$$
Now we get $z$ from (2) $$z=-frac{2 (y-12)}{y+2}$$ finally we obtain one equation in $y$ factorizing we get $$frac{7 (y-2) (y+6)}{(y+2)^2}=0$$ From here we find the Solutions of our System.
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Elimimating the variable $x$ from (1) and plug this in (3) we get
$$-frac{2 (y+4) z}{y+2}-frac{4 (y+4)}{y+2}+2 z+11=0$$
Now we get $z$ from (2) $$z=-frac{2 (y-12)}{y+2}$$ finally we obtain one equation in $y$ factorizing we get $$frac{7 (y-2) (y+6)}{(y+2)^2}=0$$ From here we find the Solutions of our System.
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 30 '18 at 19:38
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
add a comment |
add a comment |
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