Solve System of Equalities: x^2+xy+y^2, x^2+xz+z^2, y^2+yz+z^2

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In summary, x, y, and z are real and positive. equations 1), 2), and 3) are all solved in the usual way.
  • #1
solakis1
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solve the following system:

\(\displaystyle x^2 +xy +y^2=37\)
\(\displaystyle x^2+xz+z^2=28\)
\(\displaystyle y^2+yz+z^2=19\)
 
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  • #2
I presume x, y, and z are real? Are any real and positive?

-Dan
 
  • #3
yes, any reals
 
  • #4
[sp]
Here's a quick run-down.

1) \(\displaystyle x^2 + xy + y^2 = 37\)

2) \(\displaystyle x^2 + x z + z^2 = 28\)

3) \(\displaystyle y^2 + y z + z^2 = 19\)

We first need to pull a trick. For example, let's take equation 1):
\(\displaystyle x^2 + xy + y^2 = 37\)

\(\displaystyle (x - y)(x^2 + xy + y^2) = 37 (x - y)\)

\(\displaystyle x^3 - y^3 = 37 (x - y)\)

Do this to equations 2) and 3) as well.

Now add equations 1) and 3) and subtract equation 2), giving
\(\displaystyle 0 = x - 2y + z\)

Put this into (the unaltered) equations 1) and 2):
1) \(\displaystyle x^2 + xy + y^2 = 37\)

2) \(\displaystyle x^2 - 2 xy + 4y^2 = 28\)
(We can drop equation 3). It doesn't give us anything new.)

So add 2 times equation 1 and equation 2).

\(\displaystyle x^2 + 2 y^2 = 34\)

Thus
\(\displaystyle y = \pm \sqrt{ \dfrac{1}{2} (34 - x^2) }\)

Now put these y values into equation 1)
\(\displaystyle x^2 \pm x \sqrt{ \dfrac{1}{2} (34 - x^2) } + \dfrac{1}{2} (34 - x^2) = 37\)

To save some time I'll just simply say that we can solve this in the usual way of isolating the square root and squaring the whole thing. etc.

So
\(\displaystyle x = \left \{ \pm 4, ~ \pm \dfrac{10}{\sqrt{3} } \right \}\)

Putting this all together:
\(\displaystyle (x, y, z) = \begin{cases} (4, ~ 3, ~ 2) \\ (-4, ~ -3, ~ -2) \\ \left ( \dfrac{10}{\sqrt{3}}, ~ \dfrac{1}{\sqrt{3}}, ~ - \dfrac{8}{\sqrt{3}} \right ) \\ \left ( - \dfrac{10}{\sqrt{3}}, ~ - \dfrac{1}{\sqrt{3}}, ~ \dfrac{8}{\sqrt{3}} \right ) \end{cases} \)
[/sp]
-Dan
 
  • #5
topsquark said:
[sp]
Here's a quick run-down.

1) \(\displaystyle x^2 + xy + y^2 = 37\)

2) \(\displaystyle x^2 + x z + z^2 = 28\)

3) \(\displaystyle y^2 + y z + z^2 = 19\)

We first need to pull a trick. For example, let's take equation 1):
\(\displaystyle x^2 + xy + y^2 = 37\)

\(\displaystyle (x - y)(x^2 + xy + y^2) = 37 (x - y)\)

\(\displaystyle x^3 - y^3 = 37 (x - y)\)

Do this to equations 2) and 3) as well.

Now add equations 1) and 3) and subtract equation 2), giving
\(\displaystyle 0 = x - 2y + z\)

Put this into (the unaltered) equations 1) and 2):
1) \(\displaystyle x^2 + xy + y^2 = 37\)

2) \(\displaystyle x^2 - 2 xy + 4y^2 = 28\)
(We can drop equation 3). It doesn't give us anything new.)

So add 2 times equation 1 and equation 2).

\(\displaystyle x^2 + 2 y^2 = 34\)

Thus
\(\displaystyle y = \pm \sqrt{ \dfrac{1}{2} (34 - x^2) }\)

Now put these y values into equation 1)
\(\displaystyle x^2 \pm x \sqrt{ \dfrac{1}{2} (34 - x^2) } + \dfrac{1}{2} (34 - x^2) = 37\)

To save some time I'll just simply say that we can solve this in the usual way of isolating the square root and squaring the whole thing. etc.

So
\(\displaystyle x = \left \{ \pm 4, ~ \pm \dfrac{10}{\sqrt{3} } \right \}\)

Putting this all together:
\(\displaystyle (x, y, z) = \begin{cases} (4, ~ 3, ~ 2) \\ (-4, ~ -3, ~ -2) \\ \left ( \dfrac{10}{\sqrt{3}}, ~ \dfrac{1}{\sqrt{3}}, ~ - \dfrac{8}{\sqrt{3}} \right ) \\ \left ( - \dfrac{10}{\sqrt{3}}, ~ - \dfrac{1}{\sqrt{3}}, ~ \dfrac{8}{\sqrt{3}} \right ) \end{cases} \)
[/sp]
-Dan
[sp]

Subtract the 2nd equation from the 1st and the 3rd from the 2nd and we get:

(y-z)(x+y=z)=9 and (x-y)(x+y+z)=9

Divide those two and we get:

(y-z)/(x-y)=1 => z=2y-x

Substitute the above into the3rd equation and we get \(\displaystyle x^2-2xy+4y^2\) e.t.c e.t.c...[/sp]
 
  • #6
solakis said:
[sp]

Subtract the 2nd equation from the 1st and the 3rd from the 2nd and we get:

(y-z)(x+y=z)=9 and (x-y)(x+y+z)=9

Divide those two and we get:

(y-z)/(x-y)=1 => z=2y-x

Substitute the above into the3rd equation and we get \(\displaystyle x^2-2xy+4y^2\) e.t.c e.t.c...[/sp]
I'm not sure why you posted? I'm guessing that you want a verification of one of the steps. If this not what you are asking, please let me know.
[sp]
\(\displaystyle x^3 - y^3 = 37(x - y)\)

\(\displaystyle x^3 - z^3 = 28(x - z)\)

\(\displaystyle y^3 - z^3 = 19(y - z)\)

So
\(\displaystyle (x^3 - y^3) - (x^3 - z^3) + (y^3 - z^3) = (37(x - y) ) - (28(x - z)) + (19(y - z))\)

\(\displaystyle 0 = (37 - 28)x + (-37 + 19)y + (28 - 19)z\)
[/sp]

-Dan
 
  • #7
I just wanted to show that there is another way to get the equation

z=2y-x
 
  • #8
solakis said:
I just wanted to show that there is another way to get the equation

z=2y-x
Yes. That's on line 11.

-Dan
 

Related to Solve System of Equalities: x^2+xy+y^2, x^2+xz+z^2, y^2+yz+z^2

1. What is a system of equalities?

A system of equalities is a set of equations that have the same variables and are solved together to find the values of those variables that satisfy all the equations simultaneously.

2. How do you solve a system of equalities?

To solve a system of equalities, you can use various methods such as substitution, elimination, or graphing. In this case, we can use the substitution method to solve the given system of equalities.

3. What is the substitution method?

The substitution method involves solving one equation for one variable and then substituting that value into the other equation. This allows us to eliminate one variable and solve for the remaining variable.

4. How do you use the substitution method to solve this system of equalities?

To use the substitution method, we can solve the first equation, x^2+xy+y^2, for y in terms of x. This gives us y = (-x ± √(4x^2-3x^2))/2. We can then substitute this value into the second equation, x^2+xz+z^2, to get a quadratic equation in terms of x. Solving this equation will give us the values of x, which we can then substitute back into the first equation to find the corresponding values of y.

5. What is the solution to this system of equalities?

The solution to this system of equalities is the set of values for x and y that satisfy both equations. Using the substitution method, we can find that the solutions are (0,0) and (-1,1) or (1,-1).

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