1963 TMTA High School Algebra I Contest Problem 31

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Problem

Solve for $x$ when $\frac{x}{x-1} = 3 + \frac{1}{x-1}.$

$\text{(A)} \quad 2 \quad \text{(B)} \quad -1 \quad \text{(C)} \quad 1 \quad \text{(D)} \quad -2 \quad \text{(E) NOTA}$

Solution

Simplify. \[\frac{x}{x-1} = \frac{3(x-1)+1}{x-1}=\frac{3x-2}{x-1}\] Cross multiply. \[(x)(x-1)=(3x-2)(x-1)\] \[x^2 - x = 3x^2 - 5x + 2\] \[0 = 2x^2 - 4x + 2\] This yields $x = 1$ as a solution, but this is extraneous since \[1-1=0\], and the denominator in the fractions given cannot be \[0\], so the answer is $\boxed{\text{(E)}}.$

See Also

1963 TMTA High School Mathematics Contests (Problems)
Preceded by
Problem 30
TMTA High School Mathematics Contest Past Problems/Solutions Followed by
Problem 32