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Difference between revisions of "1974 AHSME Problems/Problem 1"

(Created page with "==Problem== If <math> x\not=0 </math> or <math> 4 </math> and <math> y\not=0 </math> or <math> 6 </math>, then <math> \frac{2}{x}+\frac{3}{y}=\frac{1}{2} </math> is equivalent to...")
 
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==See Also==
 
==See Also==
 
{{AHSME box|year=1974|before=First Problem|num-a=2}}
 
{{AHSME box|year=1974|before=First Problem|num-a=2}}
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[[Category:Introductory Algebra Problems]]

Revision as of 09:16, 30 May 2012

Problem

If $x\not=0$ or $4$ and $y\not=0$ or $6$, then $\frac{2}{x}+\frac{3}{y}=\frac{1}{2}$ is equivalent to

$\mathrm{(A)\ } 4x+3y=xy \qquad \mathrm{(B) \ }y=\frac{4x}{6-y} \qquad \mathrm{(C) \ } \frac{x}{2}+\frac{y}{3}=2 \qquad$

$\mathrm{(D) \ } \frac{4y}{y-6}=x \qquad \mathrm{(E) \ }\text{none of these}$

Solution

To clear out the fractions, we multiply both sides of the equation by $xy$ to get $2y+3x=\frac{xy}{2}$, or $4y+6x=xy$. From this, we have $4y=xy-6x$. We can factor an $x$ out of the RHS, and so $4y=x(y-6)\implies \frac{4y}{y-6}=x, \boxed{\text{D}}$.

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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