Efficient Solutions for Systems of 3 Equations and 3 Unknowns with Squares

[Virt]

Hi, I'm wondering what is the quickest/easiest way to solve a system of 3 equations and 3 unknowns that involves squares. Preferably not by hand. Can I use the matrix feature on my TI84, or do I need to use software like Matlab?

This set of equations came up while I was trying to solve a problem for my fluids class,

Three velocities
V1,V2,V3

.694V1+.840V2-V3=0
.373V1^2+.233V3^2=33
.296V2^2+.233V3^2=45

Thanks

 

[pwsnafu]

Wolfram Alpha knows all

http://www.wolframalpha.com/input/?i=Solve[{.694+x+%2B.840+y+-z+%3D%3D0%2C+.373+x^2%2B.233+z^2+%3D%3D+33%2C+.296+y^2%2B.233+z^2+%3D%3D45}%2C{x%2Cy%2Cz}]

Yes that is one URL. Copy and Paste into your browser.

 

[TylerH]

Make a 3x4 matrix on your calculator, enter the x, y, and z coefficients of the first equation, in that order, in the first row. In the 4th cell in that row, put the value of the equation eg., it would be 0 for your first equation, 33 for your second, etc. Do the same for rows 2 and 3. Now go to rref() in the catalog, give it the matrix as its argument, and viola, it gives you thee answers.

If your Alg II teacher didn't teach you this, you need to go yell at them, or something, If they did, but you forgot, that's more understandable.

EDIT: Oh... wait... I didn't see those squared terms. I'm not sure how you do this.

 

[BloodyFrozen]

^To my knowledge, you can't solve it using an augmented matrix.

 

[DuncanM]

The first equation has all three variables in it (v1, v2, and v3) so couldn't you use it to solve for one of them. Perhaps v3 = . . .

The second and third equations have only two of the variables in them, so use the substitution in them, and you'll have two equations in two unknowns. Hopefully, the resulting system is solvable with pen and paper. If not, the other option I can think of off the top of my head is numerically, with a multi-dimensional root-finder. (In that case, yes, you would need a software like MATLAB.)

 

[gsal]

Hhhhmmm...it took a few lines but I did but hand...

If we number the equations 1, 2, and 3 as listed, you can:

• isolate V1 from equation 2 and make it a function of V3
• isolate V2 from equation 3 and make it a function of V3
• insert the results for V1 and V2 into equation 1

...you now have a single equation with a single variable, V3.

You can solve this however you want...or can continue by hand...

Now, it is a matter of getting rid of the square roots by squaring the equality a couple of times...since I remember well how to square binomials, instead of having all 3 terms on one side of the equation equal to zero, I moved one of the radicals to the other side and square both sides of the equality (a binomial on one side and a simple term on the other).

Gather similar terms (powers of V3) and repeat the process of spliting terms to a binomial on one side and a simple term on the other and square the equality once more.

If you assign letters to constant (numeric values) terms to easy algebraic manipulation, you will end up with something like

AV₃⁴ + BV₃² + C = 0

which looks very much like a quadratic equation, so, simply substitute X = V₃²:

AX² + BX + C = 0

and there you have it, simply solve with the well known solution for a quadratic equation.

 

[NeroKid]

U can actually solve it easily with hand like this:
U check if V3 = 0 is a solution to the system or not
Now check V3 # 0
in the square equation u divide all by V3^2 and divide 1 equation withc each other
then u can get a (V1/V3)^2 = k(V2/V3)^2 + n with k and n are in R and k# 0
the first equation u change side of V3 and square them and use the square function to
reduce the V1^2 and V2^2 and then all u have is a + bv1v2 + c = v3^2
then divide it for v3 and u get the usuall 2 values 2 function

 

[HallsofIvy]

.694V1+.840V2-V3=0
.373V1^2+.233V3^2=33
.296V2^2+.233V3^2=45

can be written as a matrix equation Ax= B where
A= \begin{bmatrix}0.694 & 0.840 & -1 \\ 0.373 & 0 & 0.233 \\ 0 & 0.296 & 0.233\end{bmatrix}
x= \begin{bmatrix} V1 \\ V2\\ V3\end{bmatrix}
and
B= \begin{bmatrix} 0 \\ 45 \\ 33\end{bmatrix}

Put matrices A and B in a TI84 calculator and then x= A^{-1}B.

 

[willem2]

.694V1+.840V2-V3=0
.373V1^2+.233V3^2=33
.296V2^2+.233V3^2=45

can be written as a matrix equation Ax= B where
A= \begin{bmatrix}0.694 & 0.840 & -1 \\ 0.373 & 0 & 0.233 \\ 0 & 0.296 & 0.233\end{bmatrix}
x= \begin{bmatrix} V1 \\ V2\\ V3\end{bmatrix}
and
B= \begin{bmatrix} 0 \\ 45 \\ 33\end{bmatrix}

Put matrices A and B in a TI84 calculator and then x= A^{-1}B.

the second and third equations have the squares of v1, v2 and v3 in them, so this doesn't work. You can get a quartic equation out of it, but you're probably better of finding
a numerical solution

 

[micromass]

.694V1+.840V2-V3=0
.373V1^2+.233V3^2=33
.296V2^2+.233V3^2=45

can be written as a matrix equation Ax= B where
A= \begin{bmatrix}0.694 & 0.840 & -1 \\ 0.373 & 0 & 0.233 \\ 0 & 0.296 & 0.233\end{bmatrix}
x= \begin{bmatrix} V1 \\ V2\\ V3\end{bmatrix}
and
B= \begin{bmatrix} 0 \\ 45 \\ 33\end{bmatrix}

Put matrices A and B in a TI84 calculator and then x= A^{-1}B.

Uh, Halls, you probably missed that this is not a linear system. We have V1² and V2² in there...

 

[disregardthat]

Substitute the expression for V3 from the first equation into the others. Make two linear transformations of V1 and V2 to get rid of the coefficients of V1V2 in the remaining equations. Solve the system consisting of equations in (V1-a)^2,(V2-b)^2 by setting up a matrix.

 

[HallsofIvy]

Thanks. You are right- I missed the squares. I wondered why there was such an easy answer!