Sure! Let's solve this system of equations step-by-step.
We have the following system of equations:
1. [tex]\(\frac{1}{x+1} + \frac{1}{y-2} = -1\)[/tex]
2. [tex]\(3 \left(\frac{1}{x+1}\right)^{-2} - 2 \left(\frac{1}{y-2}\right) = 7\)[/tex]
### Step 1: Substitute [tex]\(u = \frac{1}{x+1}\)[/tex] and [tex]\(v = \frac{1}{y-2}\)[/tex]
Rewriting the equations in terms of [tex]\(u\)[/tex] and [tex]\(v\)[/tex]:
1. [tex]\(u + v = -1\)[/tex]
2. [tex]\(3u^{-2} - 2v = 7\)[/tex]
### Step 2: Solve the first equation for one variable
From the first equation:
[tex]\[ u + v = -1 \][/tex]
[tex]\[ v = -1 - u \][/tex]
### Step 3: Substitute into the second equation
Substitute [tex]\(v\)[/tex] from the first equation into the second equation:
[tex]\[ 3u^{-2} - 2(-1 - u) = 7 \][/tex]
[tex]\[ 3u^{-2} + 2 + 2u = 7 \][/tex]
[tex]\[ 3u^{-2} + 2u + 2 = 7 \][/tex]
[tex]\[ 3u^{-2} + 2u = 5 \][/tex]
### Step 4: Solve for [tex]\(u\)[/tex]
Let's find [tex]\(u\)[/tex]:
[tex]\[ 3u^{-2} + 2u - 5 = 0 \][/tex]
[tex]\[ 3 \left(\frac{1}{u^2}\right) + 2u - 5 = 0 \][/tex]
This equation might be complex to solve algebraically; we resort to numerical or symbolic solution methods. For case feasibility, let's use appropriate algebraic tools to find the value of [tex]\(u\)[/tex].
Given our final solution, we simplify the practice to focus on reasoning:
One valid solution to the equation for [tex]\(u\)[/tex] yields:
[tex]\[ u = 1 \][/tex]
If [tex]\(u = 1\)[/tex], we revisit [tex]\(v\)[/tex]:
### Step 5: Find [tex]\(v\)[/tex]
Using the first equation:
[tex]\[ u + v = -1 \][/tex]
[tex]\[ 1 + v = -1 \][/tex]
[tex]\[ v = -2 \][/tex]
### Step 6: Transform back to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
Now, recall:
[tex]\[ u = \frac{1}{x+1} \][/tex]
[tex]\[ v = \frac{1}{y-2} \][/tex]
Substituting back:
For [tex]\(u = 1\)[/tex]:
[tex]\[ 1 = \frac{1}{x+1} \][/tex]
[tex]\[ x + 1 = 1 \][/tex]
[tex]\[ x = 0 \][/tex]
For [tex]\(v = -2\)[/tex]:
[tex]\[ -2 = \frac{1}{y-2} \][/tex]
[tex]\[ y - 2 = -\frac{1}{2} \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
### Final Answer
The solution to the system of equations is:
[tex]\[ x = 0, \ y = \frac{3}{2} \][/tex]
