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Try this beautiful problem from Algebra based on Algebraic Equation.

If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

- \(5\)
- \(20\)
- \(22\)
- \(25\)
- \(36\)

algebra

Equation

sum

But try the problem first...

Answer: \(22\)

Source

Suggested Reading

AMC-10A (2001) Problem 10

Pre College Mathematics

First hint

The given equations are $xy=24$ and $xz=48$.we have to find out \(x+y+z\)

Now using two relations we can write \(\frac{xy}{xz}=\frac{24}{48}\)\(\Rightarrow 2y=z\)

Can you now finish the problem ..........

Second Hint

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get \(2y^2=72\) \( \Rightarrow y=6\) and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Final Step

Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)

- https://www.cheenta.com/integers-and-divisors-isi-b-stat-entrance-tomato-98/
- https://www.youtube.com/watch?v=9VauLfKar70

Contents

[hide]

Try this beautiful problem from Algebra based on Algebraic Equation.

If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

- \(5\)
- \(20\)
- \(22\)
- \(25\)
- \(36\)

algebra

Equation

sum

But try the problem first...

Answer: \(22\)

Source

Suggested Reading

AMC-10A (2001) Problem 10

Pre College Mathematics

First hint

The given equations are $xy=24$ and $xz=48$.we have to find out \(x+y+z\)

Now using two relations we can write \(\frac{xy}{xz}=\frac{24}{48}\)\(\Rightarrow 2y=z\)

Can you now finish the problem ..........

Second Hint

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get \(2y^2=72\) \( \Rightarrow y=6\) and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Final Step

Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)

- https://www.cheenta.com/integers-and-divisors-isi-b-stat-entrance-tomato-98/
- https://www.youtube.com/watch?v=9VauLfKar70

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