Question

how many solutions are there in a system of non linear equations

Answer 1

It depends on the equation.

Answer 2

there is no equation the answer choices are 0,1,infinite number of solutions, and any of these could be the solution

Answer 3

hmm the first answer that leaps to mind is one solution. Usually, especially in regular school math, that is the case. But not always.
1. One Solution


All linear equations ax+b=c where a is not equal to zero, have one solution:
x=c-b/a

2. No Solutions


A linear equation can have no solutions. Example: 0x + 1 = 2. Since 0x is always 0, and 0+1 = 1, we have an impossible equation 1=2. Any linear equation with no solution always has zero (0) as the coefficient before x.
3. Infinitely many solutions


A linear equation can have infinitely many solutions. Example: 0x+4=4. Since 0x is always 0 regardless of x, any value of x satisfies this equation.
Conclusions


So, be careful with problems that reduce to a linear equation. Beware that they may have no solutions or an infinite number of solutions.

Answer 4

I hope this helped :))

Answer 5

so what do u think the answer is

Answer 6

Well we'd be dealing with *non*-linear equations, so there are a lot of cases to consider.

Let's assume we're working in only two variables, so we have systems of the form
\[\begin{cases}y_1=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\\
y_2=b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0\end{cases}\]
Let's simplify the problem a bit and focus on the simplest nonlinear case: systems containing at least one quadratic, i.e.
\[(1)~\begin{cases}y=a_2x^2+a_1x+a_0\\
y=b_1x+b_0\end{cases}\]
You can have a variety of outcomes regarding solutions:

Clearly, we can't have "infinitely many solutions" unless \(a_2=0\). But then the first equation is linear, so the system is linear.

So to summarize, the answer choices you're given don't seem to work properly... Are you sure the system is nonlinear?