To solve the given system of equations:
[tex]\[
\begin{cases}
x + y = 7xy \\
2x - 3y + xy = 0
\end{cases}
\][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations.
### Step 1: Rearrange and simplify the equations
First, let's work with the first equation:
[tex]\[ x + y = 7xy \][/tex]
We can rearrange this to express it in a different form:
[tex]\[ x + y - 7xy = 0 \][/tex]
Next, let's look at the second equation:
[tex]\[ 2x - 3y + xy = 0 \][/tex]
### Step 2: Solve one equation for one variable
Let's solve the first equation for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x + y = 7xy \][/tex]
[tex]\[ y = \frac{x}{7xy - 1} \][/tex]
However, it looks simpler to attempt solving this by substitution or another method since it results in a complicated expression.
### Step 3: Consider substitution and simultaneous solution
One approach to find the solution for such equations is to consider possible values for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that could simplify the terms.
### Step 4: Check possible simple solutions
Notice that [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex] might simplify the equations dramatically:
- Substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex] in the first equation:
[tex]\[ 0 + 0 = 7 \times 0 \times 0 \][/tex]
[tex]\[ 0 = 0 \][/tex] which is true.
- Substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex] in the second equation:
[tex]\[ 2 \times 0 - 3 \times 0 + 0 \times 0 = 0 \][/tex]
[tex]\[ 0 = 0 \][/tex] which is also true.
Therefore, [tex]\((x, y) = (0, 0)\)[/tex] is one solution.
### Step 5: Find other potential solutions by solving equations simultaneously
Let's plug [tex]\( y = 1/4 \)[/tex] into the first equation to find another possible solution:
[tex]\[ x + \frac{1}{4} = 7x \left(\frac{1}{4}\right) \][/tex]
[tex]\[ x + \frac{1}{4} = \frac{7x}{4} \][/tex]
Multiplying through by 4 to clear the fraction:
[tex]\[ 4x + 1 = 7x \][/tex]
[tex]\[ 1 = 3x \][/tex]
[tex]\[ x = \frac{1}{3} \][/tex]
Check if [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( y = \frac{1}{4} \)[/tex] satisfy the second equation:
[tex]\[ 2 \times \frac{1}{3} - 3 \times \frac{1}{4} + \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = 0 \][/tex]
[tex]\[ \frac{2}{3} - \frac{3}{4} + \frac{1}{12} = 0 \][/tex]
Convert to a common denominator (12):
[tex]\[ \frac{8}{12} - \frac{9}{12} + \frac{1}{12} = 0 \][/tex]
[tex]\[ \frac{8 - 9 + 1}{12} = 0 \][/tex]
[tex]\[ \frac{0}{12} = 0 \][/tex]
This is also true, so [tex]\((x, y) = \left(\frac{1}{3}, \frac{1}{4}\right)\)[/tex] is another solution.
### Conclusion:
The solutions to the system of equations are:
[tex]\[ \boxed{(0, 0) \text{ and } \left(\frac{1}{3}, \frac{1}{4}\right)} \][/tex]
