How to find 3 unknowns with 2 simultaneous equations?

Answer 1

You can't really find a real answer as you don't have enough equations.

You can get a parametrized answer. For example:

2x+y+z=1
z+x = 2

We can't have a real answer, so we set the value of one of the variables to be a parameter.
x=t
And this is used as a 'third equation'
eventually we get:
x = t
y = -(t+1)
z = 2-t

Answer 2

You point out us that the answer is : (a = 6, b = - 6, c = a million) in case you want this answer to be wonderful, i think of there's slightly mistake in the equation (3). you've gotten : 9a + 3b + c = fifty 4 - 18 + a million = 37 instead of 36 (a million) : a + b + c = a million (2) : 4a + 2b + c = 13 (3) : 9a + 3b + c = 37 Do the calculation : (2) - (a million), and get equation (4) (4) : 4a + 2b + c - a - b - c = 13 - a million (4) : 3a + b = 12 (4) : b = 12 - 3a in accordance the equation (3) (3) : c = 37 - 9a - 3b replace "b" and "c" by them values into the equation (a million) and get equation (5) : (5) : a + [b] + [c] = a million (5) : a + [12 - 3a] + [37 - 9a - 3b] = a million (5) : a + 12 - 3a + 37 - 9a - 3b = a million (5) : - 11a - 3b = a million - 12 - 37 (5) : - 11a - 3b = - 40 8 (5) : 11a + 3b = 40 8 From equation (4) : 3a + b = 12, multiply by 3 and get equation (6) : (6) : 9a + 3b = 36 Now, calculate (5) - (6) and get equation (7) : (7) : 11a + 3b - 9a - 3b = 40 8 - 36 (7) : 2a = 12 (7) : a = 6 in accordance (4) : b = 12 - 3a b = 12 - 3(6) b = 12 - 18 b = - 6 in accordance (a million) : a + b + c = a million c = a million - a - b c = a million - (6) - (- 6) c = a million - 6 + 6 c = a million Finaly, you acquire : a = 6, b = - 6, c = a million

Answer 3

Principally it is not possible. But there are ways out as follows. 1)Then there is a process called Gaussian elimination which might do the trick, provided one of the terms can be infered from the equations. This is done by making last term of one equation equal to last term of the other and subtracting. Then you have two equation and two unknowns. Now you make the last term of this new equation equal to the last term of the first equation and subtract then you get one term. Now keep on substituting this in the other equations got and you have all the answers.
2)Simplex method in Linear Programming is another way out. 3)If you say you have a 2x3 matrix you multiply the matrix by it's transpose which is 3x2 and you will get a 3x3 matrix which is three equations and three unknwons. Now solving is easy.

Answer 4

Elementary Mandy..... check it out.


Solve this system of three equations in three unknowns:

1) x + y − z = 4

2) x − 2y + 3z = −6

3) 2x + 3y + z = 7

The strategy is to reduce this to two equations in two unknowns. Do this by eliminating one of the unknowns from two pairs of equations: either from equations 1) and 2), or 1) and 3), or 2) and 3).

For example, let us eliminate z. We will first eliminate it from equations 1) and 3) simply by adding them. We obtain:

4) 3x + 4y = 11

Next, we will eliminate z from equations 1) and 2). We will multiply equation 1) by 3. We call the resulting equation 1' ("1 prime") to show that we obtained it from equation 1):

1') 3x + 3y − 3z = 12

2) x − 2y + 3z = −6
______________________________________________________________________________________
5) 4x + y = 6

We now solve equations 4) and 5) for x and y.

Let us eliminate y. We will multiply equation 5) by −4, and add it to equation 4):

5') −16x − 4y = −24

4) 3x + 4y = 11
____________________________________________________
−13x = −13

x = 1

To solve for y, let us substitute x = 1 in equation 4):

3 + 4y = 11

4y = 11 −3

4y = 8

y = 2

Finally, to solve for z, substitute these values of x and y in one of the original equations; say equation 1):

1 + 2 − z = 4

−z = 4 − 3 = 1

z = −1

Answer 5

Put the 2 equations together and you'll get your 3 unknowns. LOL

Answer 6

there cannot be one ans.

either there shall be zero ans or one of them can take any arbritrary value and other 2 can be expressed in terms of them

for example

x+ y + z= 13
2x + 2y + 2z = 27
the above has 0 ans

x+y + z = 13 ...1
x+2y + z = 17... 2
we can subraact 1st from 2nd get
y= 4
from 1
x+ z= 9
from 2 x +z = 9
so
(x, y= 4, z = 9-x) is a solution x is arbitrary
y cannot be arbritrary

Answer 7

I'd reccommend finding a relationship between the unknowns and just use one equation with one or two variables.

Answer 8

Not possible ( unless a trivial case all equal zero ), to fully solve a problem you need a as many equations as unknows

Answer 9

Try 'grouping' the two unknowns and making them one e.g.:
y+2x+z=3
2y+5x+2z=8

group y and z so you can find x and then when you have found it just solve the simult equs normaly with two unknowns [y, z]

the answer to this would be:
2y+4x+2z=6
2y+5x+2z=8
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now minus both sides and you get
-x=-2
x=2

then u got x so now substitute it into the equs and voila! u gt ur usual simult. equs.

Answer 10

I have no idea.