Tessellate S.T.E.M.S (2019) - Mathematics - Category B - Set 3 - Objective Problem 2

$x+\frac { 1 }{ z } =5$ [Given]

$y+\frac { 1 }{ x } =29$ [Given]

Let $z+\frac { 1 }{ y } =A$

Multiplying the three equations gives us:

$xyz+\frac { 1 }{ xyz } +x+y+z+\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } =145A$

Rearranging:

$\left( xyz+\frac { 1 }{ xyz } \right) + \left( x+\frac { 1 }{ z } \right) +\left( y+\frac { 1 }{ x } \right)+\left( z+\frac { 1 }{ y } \right) =145A$

Substituting $xyz=1$ , $x+\frac { 1 }{ z } =5$ , $y+\frac { 1 }{ x } =29$ [All given], and $z+\frac { 1 }{ y } =A$ [Our earlier assumption]:

$\left( 1+\frac { 1 }{ 1 } \right) +5+29+A=145A$

Simplifying:

$36+A=145A$

Solving for A:

$A=\frac { 1 }{ 4 }$

Remember that our assumption is:

$A=z+\frac { 1 }{ y }$ which is said to equal $\frac { m }{ n }$

Thus, $\frac { m }{ n } =\frac { 1 }{ 4 }$

Therefore, $m=1$ and $n=4$

Since we are asked for $m+n$ , the answer is therefore $1+4=\boxed { 5 }$

$xz+1=5z$

$\frac {1}{y}+1=5z$

$1+y=5zy=\frac {5}{x}$ , $y+\frac {1}{x}=29$

Two equations, two variables ( $y$ and $\frac {1}{x}$ )

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$y=24, \frac {1}{x}=5$

Put values in first equation to get $z=\frac {5}{24}$

$z+\frac {1}{y}=\frac {1}{4}$