Haynes Mills
01/20/2023 · Primary School
\( \frac{(x-3)^{2}}{x-2}+\frac{15}{3}=\frac{x^{2}-x-1}{x-2} \quad \) [ind., \( \left.x \neq 2\right] \)
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Tutor-Verified Answer
Step-by-step Solution
Solve the equation \( \frac{(x-3)^{2}}{x-2}+\frac{15}{3}=\frac{x^{2}-x-1}{x-2} \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{\left(x-3\right)^{2}}{x-2}+\frac{15}{3}=\frac{x^{2}-x-1}{x-2}\)
- step1: Find the domain:
\(\frac{\left(x-3\right)^{2}}{x-2}+\frac{15}{3}=\frac{x^{2}-x-1}{x-2},x\neq 2\)
- step2: Divide the terms:
\(\frac{\left(x-3\right)^{2}}{x-2}+5=\frac{x^{2}-x-1}{x-2}\)
- step3: Multiply both sides of the equation by LCD:
\(\left(\frac{\left(x-3\right)^{2}}{x-2}+5\right)\left(x-2\right)=\frac{x^{2}-x-1}{x-2}\times \left(x-2\right)\)
- step4: Simplify the equation:
\(x^{2}-x-1=x^{2}-x-1\)
- step5: The statement is true:
\(x \in \mathbb{R}\)
- step6: Check if the solution is in the defined range:
\(x \in \mathbb{R},x\neq 2\)
- step7: Find the intersection:
\(x\neq 2\)
The solution to the given equation is \( x \neq 2 \).
Quick Answer
The solution to the equation is \( x \neq 2 \).
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