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# Algebraic Equation | AIME I, 2000 Question 7 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation - AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, $x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$ then $z+\frac{1}{y}$=$\frac{m}{n}$ where m and n are relatively prime, find m+n

• is 107
• is 5
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Integers

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here $x+\frac{1}{z}=5$ then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and $y=(29-\frac{1}{x}$) together gives 5-x=x$(29-\frac{1}{x}$) then x=$\frac{1}{5}$

then y=29-5=24 and z=$\frac{1}{5-x}$=$\frac{5}{24}$

$z+\frac{1}{y}$=$\frac{1}{4}$ then 1+4=5.

.

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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation - AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, $x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$ then $z+\frac{1}{y}$=$\frac{m}{n}$ where m and n are relatively prime, find m+n

• is 107
• is 5
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Integers

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here $x+\frac{1}{z}=5$ then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and $y=(29-\frac{1}{x}$) together gives 5-x=x$(29-\frac{1}{x}$) then x=$\frac{1}{5}$

then y=29-5=24 and z=$\frac{1}{5-x}$=$\frac{5}{24}$

$z+\frac{1}{y}$=$\frac{1}{4}$ then 1+4=5.

.

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