can we solve 3 equations having 3 unknowns of 3 different degrees,?if not ,why?

Question

do u think u r a genius in maths? then try this

Answer 1

You can.
First take 1st & 2nd equations & eliminate 1 variable(unknown) by simultaneous equation.The take 2nd & 3rd equations & eliminate the same unknown as you did in 1st case. Now you have 2 equations with 2 unknown . Apply simultaneous equations to find the 2 unknows , the you can find the 3rd unknown easily.

Answer 2

Yes we can solve 3 equations having 3 unknowns of 3 different degrees.
Because theory of equation does not have any confinement of degree, Theory of Equation States:

In mathematics, equation solving is the problem of finding what values (numbers, functions, sets, etc.) fulfill a condition stated as an equality (an equation). Usually, this condition involves expressions with variables (or unknowns), which are to be substituted by values in order for the equality to hold. More precisely, an equation involves some free variables.

In one general case, we have a situation such as

f(x1,...,xn)=c,
with c being a constant, which has a set of solutions S of the form

{(a1,...,an)∈Tn|f(a0,...,an)=c}
with Tn the domain of the function. Note that the set of solutions can be empty (there are no solutions), singleton (there is exactly 1 solution), finite, or infinite (there are infinitely many solutions).

For example, an expression such as

3x+2y=21z
can be solved by first modifying the equation in some way as to preserve the equality, such as subtracting both sides by 21z to obtain

3x+2y-21z=0
Now, it occurs that in solving this equation, that there is not just one solution to this equation, but an infinite set of solutions, which can be written

{(x, y, z)|3x+2y-21z=0}.
One particular solution is x = 20/3, y = 11, z = 2. In fact, this particular set of solutions describe a plane in three dimensions, which passes through the point (20/3, 11, 2).

Answer 3

If equations are having 3 variables and 3 LINEAR equations it might be possible to solve them. But if they are having different degrees then it requires lot of reasearch. By different degrees you could mean, any starting from 1,2,3...1000.... If you are okay with imaginary solutions, then I can think further. I MAY NOT BE A GENIUS, BUT I AM IN THE MAKING!!!

Answer 4

I think yes. 3 equations are sufficient for solving 3 unknowns.

Answer 5

im not 100% sure but i dont think the degrees matter. if you have X squared thats just one unknown .. same as having X. so 3 equations to solve 3 unknowns shouldn be a problem.

Answer 6

there isnt extremely an person-friendly technique, except you've a calculator. positioned right into a 3x4 matrix and do rref, or positioned coefficients in a 3x3(A) matrix and the constants in a 3x1(B) and then do A^-a million*B i imagine a thanks to do those is gaussian technique. really upload and subtract rows till you get into rref form. 2x - y - z = 17 (a million) x + 3y + 4z = -20 (2) 5x - 2y + 3z = 19 (3) (a million)-2(2) into (2) and 5(a million)-2(3) into (3) 2x - y - z = 17 (a million) -7y - 9z = fifty seven (2) -y - 11z = 40 seven (3) (2) - 7(3) into (3) and (a million)-(3) into (a million) 2x + 10z = -30 (a million) -7y - 9z = fifty seven (2) 68z = -272 (3) divide (3)/sixty 8 then 9(3)+(2) into (2) and (a million)-10(3) 2x = 10 (a million) -7y = 21 (2) z = -4 (3) do (a million)/2 and (2)/-7 and think with reference to the answer x = 5 y = -3 z = -4 make it a sturdy day

Answer 7

yes
you can use trial and error method to solve them
or you may use a computer program to check the equalities for all values of unknowns.
But in analitical way, in general, it is difficult to solve.

Answer 8

yes of course

Answer 9

Of course!!

I didn't choose no, because I did not want to explain myself : )