Sure, let's solve the system of equations step-by-step.
The given system of equations is:
1. [tex]\(4x + 9y - 5xy = 0\)[/tex]
2. [tex]\(\frac{x + 2y}{xy} = 2\)[/tex]
First, let's rewrite the second equation to make it easier to handle algebraically:
[tex]\[
\frac{x + 2y}{xy} = 2
\][/tex]
Multiplying both sides by [tex]\(xy\)[/tex] to clear the fraction, we get:
[tex]\[
x + 2y = 2xy
\][/tex]
Rearranging this equation, we get:
[tex]\[
x + 2y - 2xy = 0 \quad \text{(Equation 3)}
\][/tex]
Now we have the following system of equations:
1. [tex]\(4x + 9y - 5xy = 0\)[/tex]
2. [tex]\(x + 2y - 2xy = 0\)[/tex]
Next, let's solve the second equation for one of the variables, say [tex]\(x\)[/tex]:
From Equation 3:
[tex]\[
x + 2y - 2xy = 0
\][/tex]
Factoring [tex]\(x\)[/tex] out from the terms involving [tex]\(x\)[/tex], we get:
[tex]\[
x(1 - 2y) = -2y
\][/tex]
Assuming [tex]\(1 - 2y \neq 0\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-2y}{1 - 2y}
\][/tex]
Now substitute [tex]\(x\)[/tex] from this expression into the first equation.
Substitute [tex]\(x = \frac{-2y}{1 - 2y}\)[/tex] into Equation 1:
[tex]\[
4 \left(\frac{-2y}{1 - 2y}\right) + 9y - 5 \left(\frac{-2y}{1 - 2y}\right)y = 0
\][/tex]
Simplifying each term separately:
[tex]\[
4 \left(\frac{-2y}{1 - 2y}\right) = \frac{-8y}{1 - 2y}
\][/tex]
[tex]\[
-5 \left(\frac{-2y}{1 - 2y}\right)y = \frac{10y^2}{1 - 2y}
\][/tex]
Thus the equation becomes:
[tex]\[
\frac{-8y + 9y(1 - 2y) + 10y^2}{1 - 2y} = 0
\][/tex]
Simplifying the numerator:
[tex]\[
-8y + 9y - 18y^2 + 10y^2 = 0
\][/tex]
Combining like terms:
[tex]\[
y - 8y^2 = 0
\][/tex]
Factoring out [tex]\(y\)[/tex]:
[tex]\[
y(1 - 8y) = 0
\][/tex]
This gives us two possible solutions for [tex]\(y\)[/tex]:
[tex]\[
y = 0 \quad \text{or} \quad 1 - 8y = 0
\][/tex]
Solving [tex]\(1 - 8y = 0\)[/tex], we get:
[tex]\[
y = \frac{1}{8}
\][/tex]
For [tex]\(y = 0\)[/tex], substituting in the second equation:
[tex]\[
x + 2(0) - 2x(0) = 0 \implies x = 0
\][/tex]
However, [tex]\(x = 0\)[/tex] does not satisfy the first equation, so this solution is discarded.
For [tex]\(y = \frac{1}{8}\)[/tex]:
Substituting [tex]\(y = \frac{1}{8}\)[/tex] back into [tex]\(x = \frac{-2y}{1 - 2y}\)[/tex]:
[tex]\[
x = \frac{-2(\frac{1}{8})}{1 - 2(\frac{1}{8})} = \frac{-\frac{1}{4}}{1 - \frac{1}{4}} = \frac{-\frac{1}{4}}{\frac{3}{4}} = -\frac{1}{3}
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
x = -\frac{1}{3} \quad \text{and} \quad y = \frac{1}{8}
\][/tex]
So the solution is [tex]\((- \frac{1}{3}, \frac{1}{8})\)[/tex].
