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If x, y, z are positive integers, such that xy + x + y = 215, yz + y + z = 161 and zx + z + x = 107, then (y – z)/(x – z) =

Question

If x, y, z are positive integers, such that

  • xy+x+y=215 xy + x + y = 215
  • yz+y+z=161 yz + y + z = 161
  • zx+z+x=107 zx + z + x = 107

then (yz)(xz)= \frac{(y – z)}{(x – z)} =

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Solution

1. Break Down the Problem

We have three equations involving positive integers xx, yy, and zz:

  1. xy+x+y=215xy + x + y = 215
  2. yz+y+z=161yz + y + z = 161
  3. zx+z+x=107zx + z + x = 107

We need to find the value of yzxz\frac{y - z}{x - z}.

2. Relevant Concepts

We can rearrange the equations to express them in a more manageable form:

  • From xy+x+y=215xy + x + y = 215, we can rearrange to xy+x+y215=0xy + x + y - 215 = 0.
  • Similar rearrangements can be made for the other two equations.

Let's rewrite each equation:

  • xy+x+y=215    (x+1)(y+1)=216xy + x + y = 215 \implies (x + 1)(y + 1) = 216
  • yz+y+z=161    (y+1)(z+1)=162yz + y + z = 161 \implies (y + 1)(z + 1) = 162
  • zx+z+x=107    (z+1)(x+1)=108zx + z + x = 107 \implies (z + 1)(x + 1) = 108

3. Analysis and Detail

Now we will factor each equation based on the products obtained:

  1. First Equation: (x+1)(y+1)=216(x + 1)(y + 1) = 216 The factors of 216 are: 1×216,2×108,3×72,4×54,6×36,8×27,9×24,12×18 1 \times 216, 2 \times 108, 3 \times 72, 4 \times 54, 6 \times 36, 8 \times 27, 9 \times 24, 12 \times 18

  2. Second Equation: (y+1)(z+1)=162(y + 1)(z + 1) = 162 The factors of 162 are: 1×162,2×81,3×54,6×27,9×18,18×9 1 \times 162, 2 \times 81, 3 \times 54, 6 \times 27, 9 \times 18, 18 \times 9

  3. Third Equation: (z+1)(x+1)=108(z + 1)(x + 1) = 108 The factors of 108 are: 1×108,2×54,3×36,4×27,6×18,9×12 1 \times 108, 2 \times 54, 3 \times 36, 4 \times 27, 6 \times 18, 9 \times 12

We will find valid combinations (x+1,y+1,z+1) (x + 1, y + 1, z + 1) from the factor pairs of each equation.

4. Verification through substitution

Let’s try combining factors that appear among the products:

Starting with the equation (x+1)(y+1)=216(x + 1)(y + 1) = 216:

  • Let x+1=9x + 1 = 9 and y+1=24y + 1 = 24, then x=8x = 8 and y=23y = 23.

Next, substituting into (y+1)(z+1)=162(y + 1)(z + 1) = 162:

  • If y+1=24y + 1 = 24, then z+1=6.75z + 1 = 6.75 (not valid).

So we discard this pair and try another:

Set x+1=12x + 1 = 12 (then x=11x = 11) and y+1=18y + 1 = 18 (then y=17y = 17):

  • Calculate for zz using (y+1)(z+1)=162(y + 1)(z + 1) = 162: 18(z+1)=162    z+1=9    z=8 18(z + 1) = 162 \implies z + 1 = 9 \implies z = 8

Now we verify with the third equation:

  • Check (z+1)(x+1)=108(z + 1)(x + 1) = 108: 9×12=108(valid) 9 \times 12 = 108 \quad \text{(valid)}

Final Answer

Now we can compute: yzxz=178118=93=3 \frac{y - z}{x - z} = \frac{17 - 8}{11 - 8} = \frac{9}{3} = 3

Thus, the final answer is: 3 \boxed{3}

This problem has been solved

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