Question

topic: identifying extraneous solutions

1. below is the work done to solve a rational equation. the problem has been worked correctly. explain why the equation has only one solution.

solve: (\frac{2}{x^{2}-2x}-\frac{1}{x - 2}=1)
(\frac{2}{x(x - 2)}-\frac{(x)1}{(x)(x - 2)} = 1) (longleftarrow) write using a common denominator.
(\frac{2 - x}{(x)(x - 2)} = 1) (longleftarrow) subtract.
((x)(x - 2)\frac{2 - x}{(x)(x - 2)} = 1(x)(x - 2)) (longleftarrow) multiply both sides by the common denominator.
(2 - x = x^{2}-2x) (longleftarrow) simplify.
(x^{2}-x - 2 = 0) (longleftarrow) write a quadratic equation in standard form.
((x - 2)(x + 1)=0) (longleftarrow) factor
(x = 2) or (x=-1) (longleftarrow) apply the zero - product property and solve for (x)
(longleftarrow) substitute 2 and - 1 into the original equation to see if the numbers are solutions.
substitute the given numbers into the given equation. identify which are actual solutions and which, if any, are extraneous.

1. (a:-1) and (\frac{5}{2}) (a-\frac{3}{2a + 1}=2)
2. (d:0) and 3 (\frac{3d}{d^{2}-d}-\frac{1}{d - 1}=1)
3. (m:1) (\frac{1}{m^{2}-m}-\frac{1}{m - 1}=0)

solve 5 and 6.
watch for extraneous solutions.

1. (\frac{1}{x^{2}-x}-\frac{1}{x - 1}=\frac{1}{2})
2. (2x+\frac{3}{x + 2}=1)

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Explanation:

Problem 1: Analyzing the Rational Equation Solution

Step 1: Check Domain Restrictions

The original equation is \(\frac{2}{x^2 - 2x}-\frac{1}{x - 2}=1\). Factor the denominator of the first term: \(x^2-2x=x(x - 2)\). So the domain excludes \(x = 0\) and \(x=2\) (since these make denominators zero).

Step 2: Substitute \(x = 2\)

Substitute \(x = 2\) into the original equation. The denominators \(x^2-2x=2^2 - 2\times2=0\) and \(x - 2=0\), so \(x = 2\) is not in the domain. Thus, it's extraneous.

Step 3: Substitute \(x=-1\)

Substitute \(x=-1\) into the original equation:
Left - hand side (LHS): \(\frac{2}{(-1)^2-2\times(-1)}-\frac{1}{-1 - 2}=\frac{2}{1 + 2}-\frac{1}{-3}=\frac{2}{3}+\frac{1}{3}=1\), which equals the right - hand side (RHS) of the equation. So \(x=-1\) is a valid solution.

So the equation has only one solution (\(x=-1\)) because \(x = 2\) is extraneous (not in the domain of the original equation).

Problem 2a: Checking \(a=-1\) and \(a = \frac{5}{2}\) for \(a-\frac{3}{2a + 1}=2\)

Step 1: Check Domain Restriction

The denominator \(2a + 1
eq0\Rightarrow a
eq-\frac{1}{2}\). Both \(a=-1\) and \(a=\frac{5}{2}\) are in the domain.

Step 2: Substitute \(a=-1\)

LHS: \(-1-\frac{3}{2\times(-1)+1}=-1-\frac{3}{-2 + 1}=-1+3 = 2\), which equals RHS. So \(a=-1\) is a solution.

Step 3: Substitute \(a=\frac{5}{2}\)

LHS: \(\frac{5}{2}-\frac{3}{2\times\frac{5}{2}+1}=\frac{5}{2}-\frac{3}{5 + 1}=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2\), which equals RHS. So \(a=\frac{5}{2}\) is a solution.

Problem 3: Checking \(d = 0\) and \(d = 3\) for \(\frac{3d}{d^2 - d}-\frac{1}{d - 1}=1\)

Step 1: Check Domain Restriction

Factor \(d^2 - d=d(d - 1)\), so domain excludes \(d = 0\) and \(d = 1\).

Step 2: Substitute \(d = 0\)

The denominator \(d^2 - d=0\) and \(d=0\) makes the first term's denominator zero, so \(d = 0\) is extraneous.

Step 3: Substitute \(d = 3\)

LHS: \(\frac{3\times3}{3^2-3}-\frac{1}{3 - 1}=\frac{9}{9 - 3}-\frac{1}{2}=\frac{9}{6}-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}=1\), which equals RHS. So \(d = 3\) is a solution, \(d = 0\) is extraneous.

Problem 4: Checking \(m = 1\) for \(\frac{1}{m^2 - m}-\frac{1}{m - 1}=0\)

Step 1: Check Domain Restriction

Factor \(m^2 - m=m(m - 1)\), so domain excludes \(m = 0\) and \(m = 1\).

Step 2: Substitute \(m = 1\)

The denominators \(m^2 - m=0\) and \(m - 1=0\), so \(m = 1\) is extraneous (no solution from \(m = 1\)).

Problem 5: Solve \(\frac{1}{x^2 - x}-\frac{1}{x - 1}=\frac{1}{2}\)

Step 1: Factor Denominators and Find Domain

Factor \(x^2 - x=x(x - 1)\). Domain: \(x
eq0\) and \(x
eq1\).

Step 2: Get Common Denominator

The common denominator is \(x(x - 1)\). Rewrite the equation: \(\frac{1}{x(x - 1)}-\frac{x}{x(x - 1)}=\frac{1}{2}\)

Step 3: Subtract the Fractions

\(\frac{1 - x}{x(x - 1)}=\frac{1}{2}\). Notice that \(1 - x=-(x - 1)\), so \(\frac{-(x - 1)}{x(x - 1)}=\frac{1}{2}\). Cancel \(x - 1\) (valid since \(x
eq1\)): \(\frac{-1}{x}=\frac{1}{2}\)

Step 4: Solve for \(x\)

Cross - multiply: \(x=-2\)

Step 5: Check Solution

Substitute \(x=-2\) into the original equation:
LHS: \(\frac{1}{(-2)^2-(-2)}-\frac{1}{-2 - 1}=\frac{1}{4 + 2}+\frac{1}{3}=\frac{1}{6}+\frac{1}{3}=\frac{1 + 2}{6}=\frac{1}{2}\), which equals RHS. And \(x=-2\) is in the domain. So the solution is \(x=-2\).

Problem 6: Solve \(2x+\frac{3}{x + 2}=1\)

Step 1: Find Domain

Domain: \(x
eq - 2\)

Step 2: Multiply by Common Denominator

Multiply both sides by \(x + 2\): \(2x(x + 2)+3=(x + 2)\)

Step 3: Expand and Simplify

Expand: \(2x^2+4x + 3=x + 2\)
Rearrange to standard quadratic form: \(2x^2+3x + 1=0\)

Step…

Answer:

s:

1. The equation has only one solution (\(x=-1\)) because \(x = 2\) is extraneous (not in the domain).
2. \(a=-1\) and \(a=\frac{5}{2}\) are both solutions.
3. \(d = 3\) is a solution, \(d = 0\) is extraneous.
4. \(m = 1\) is extraneous (no solution from \(m = 1\)).
5. The solution is \(x=-2\).
6. The solutions are \(x=-\frac{1}{2}\) and \(x=-1\).