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Solve the following system of equations (1)

Solve the following system of equationsHow to solve this system of equations.Solve this system by rewriting in row-echelon form $x+y+z=6$, $2x-y+z=3$, $3x-z=0$solve a system of equations by matricesThe system of Diophantine equations.Solve system of trigonometric equationsNonlinear system Diophantus.How to solve system of equations involving square rootsSolve the system of nonlinear equationsIdeas to solve the following system of equations

0

1

$begingroup$

Solve the following system of equations:
$$large
left {
beginaligned
sqrtx + 2(x - y + 3) &= sqrty\
x^2 + (x + 3)(2x - y + 5) &= x + 16
endaligned
right.$$

That is definitely not easy to solve for me.
I try to solve the question by letting $x + 2 = a$ and $x - y + 3 = b$ but it didn't work.

polynomials systems-of-equations

share|cite|improve this question

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

$endgroup$

add a comment | 

0

1

$begingroup$

Solve the following system of equations:
$$large
left {
beginaligned
sqrtx + 2(x - y + 3) &= sqrty\
x^2 + (x + 3)(2x - y + 5) &= x + 16
endaligned
right.$$

That is definitely not easy to solve for me.
I try to solve the question by letting $x + 2 = a$ and $x - y + 3 = b$ but it didn't work.

polynomials systems-of-equations

share|cite|improve this question

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

$endgroup$

add a comment | 

$begingroup$

Solve the following system of equations:
$$large
left {
beginaligned
sqrtx + 2(x - y + 3) &= sqrty\
x^2 + (x + 3)(2x - y + 5) &= x + 16
endaligned
right.$$

That is definitely not easy to solve for me.
I try to solve the question by letting $x + 2 = a$ and $x - y + 3 = b$ but it didn't work.

polynomials systems-of-equations

share|cite|improve this question

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

$endgroup$

share|cite|improve this question

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

share|cite|improve this question

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

edited Mar 17 at 14:17

Maria Mazur

48.5k1260121

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

asked Mar 17 at 13:59

Lê Thành ĐạtLê Thành Đạt

30611

2 Answers
2

active

oldest

votes

3

$begingroup$

Let $a=sqrtx+2$ and $b=sqrty$, then first equation is $$a^3+a-ab^2=b$$ so $$ a(a-b)(a+b)+a-b=0$$
and thus $$(a-b)(a^2+ab+1)=0$$

Case 1: $a=b$ so $y=x+2$:...

Case 2: $a^2+ab =-1$ is impossible since $a,bgeq 0$.

share|cite|improve this answer

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

$endgroup$

add a comment | 

1

$begingroup$

Hint: Squaring the first equation and factorizing we get
$$(x-y+2)(x^2-xy+6x-2y+9)=0$$ so we get
$$x-y+2=0$$
or
$$x^2-xy+6x-2y+9=0$$
Can you proceed now?

share|cite|improve this answer

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where is the other system?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 20 at 15:48
  

  
  
  
  

add a comment | 

Your Answer

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3

$begingroup$

Let $a=sqrtx+2$ and $b=sqrty$, then first equation is $$a^3+a-ab^2=b$$ so $$ a(a-b)(a+b)+a-b=0$$
and thus $$(a-b)(a^2+ab+1)=0$$

Case 1: $a=b$ so $y=x+2$:...

Case 2: $a^2+ab =-1$ is impossible since $a,bgeq 0$.

share|cite|improve this answer

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

$endgroup$

add a comment | 

3

$begingroup$

Let $a=sqrtx+2$ and $b=sqrty$, then first equation is $$a^3+a-ab^2=b$$ so $$ a(a-b)(a+b)+a-b=0$$
and thus $$(a-b)(a^2+ab+1)=0$$

Case 1: $a=b$ so $y=x+2$:...

Case 2: $a^2+ab =-1$ is impossible since $a,bgeq 0$.

share|cite|improve this answer

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

$endgroup$

add a comment | 

$begingroup$

Let $a=sqrtx+2$ and $b=sqrty$, then first equation is $$a^3+a-ab^2=b$$ so $$ a(a-b)(a+b)+a-b=0$$
and thus $$(a-b)(a^2+ab+1)=0$$

Case 1: $a=b$ so $y=x+2$:...

Case 2: $a^2+ab =-1$ is impossible since $a,bgeq 0$.

share|cite|improve this answer

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

$endgroup$

share|cite|improve this answer

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

answered Mar 17 at 14:13

Maria MazurMaria Mazur

48.5k1260121

1

$begingroup$

Hint: Squaring the first equation and factorizing we get
$$(x-y+2)(x^2-xy+6x-2y+9)=0$$ so we get
$$x-y+2=0$$
or
$$x^2-xy+6x-2y+9=0$$
Can you proceed now?

share|cite|improve this answer

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where is the other system?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 20 at 15:48
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: Squaring the first equation and factorizing we get
$$(x-y+2)(x^2-xy+6x-2y+9)=0$$ so we get
$$x-y+2=0$$
or
$$x^2-xy+6x-2y+9=0$$
Can you proceed now?

share|cite|improve this answer

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where is the other system?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 20 at 15:48
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Squaring the first equation and factorizing we get
$$(x-y+2)(x^2-xy+6x-2y+9)=0$$ so we get
$$x-y+2=0$$
or
$$x^2-xy+6x-2y+9=0$$
Can you proceed now?

share|cite|improve this answer

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

$endgroup$

share|cite|improve this answer

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867

answered Mar 17 at 14:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.1k42867