\(x + y + xy = 19, \quad y + z + yz = 29, \quad z + x + zx = 23\)
If x, y, and z are positive real numbers satisfying the system above, then find x, y, and z.
+23246 x + y + xy = 19 ---> y + xy = 19 - x ---> y(1 + x) = 19 - x
---> y = (19 - x) / (x + 1) [Equation #1]
y + z + yz = 29 ---> y + yz = 29 - z ---> y(1 + z) = 29 - z
---> y = (29 - z) / (z + 1) [Equation #2]
z + x + xz = 23 ---> x + zx = 23 - z ---> x(1 + z) = 23 - z
---> x = (23 - z) / (z + 1) [Equation #3]
Substituting Equation #3 into Equations #1:
y = (19 - x) / (x + 1) ---> y = [ 19 - (23 - z) / (z + 1) ] / [ (23 - z) / (z + 1) + 1 ]
Multiplying numerator and denominator by (z + 1)
---> y = [ 19(z + 1) - (23 - z) ] / [ (23 - z) + 1(z + 1) ]
---> y = [ 19z + 19 - 23 + z ] / [ 23 - z + z + 1 ]
---> y = [ 20z - 4 ] / [ 24 ]
---> y = (5z - 1) / 6 [Equation #4]
Setting Equation #2 and Equation #4 equal to each other:
(29 - z) / (z + 1) = (5z - 1) / 6
Cross-multiplying: 6(29 - z) = (5z - 1)(z + 1)
174 - 6z = 5z2 + 5z - z + 1
0 = z2 + 2z - 35
0 = (z + 7)(z - 5)
So: either z = -7 or z = 5 (the problem specifies that z must be positive ---> z = 5)
Equation #2: y = (29 - z) / (z + 1) ---> y = (29 - z) / (5 + 1) ---> y = 4
Equation #3: x = (23 - z) / (z + 1) ---> x = (23 - 5) / (5 + 1) ---> x = 3
+26367 If x, y, and z are positive real numbers satisfying the system above, then find x, y, and z.
\(x + y + xy = 19, \quad y + z + yz = 29, \quad z + x + zx = 23\)
\(\begin{array}{|rcll|} \hline \mathbf{x + y + xy} &=& \mathbf{19} \\ x (1+y)+ y &=& 19 \\ x (1+y)+ y+1-1 &=& 19 \\ x (1+y)+ (1+y) &=& 20 \\ \mathbf{(1+y)(1+x)} &=& \mathbf{20}\quad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{y + z + yz } &=& \mathbf{29} \\ y (1+z)+ z &=& 29 \\ y (1+z)+ z+1-1 &=& 29 \\ y (1+z)+ (1+z) &=& 30 \\ \mathbf{(1+z)(1+y)} &=& \mathbf{30}\quad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{z + x + zx } &=& \mathbf{23} \\ z (1+x)+ x &=& 23 \\ z (1+x)+ x+1-1 &=& 23 \\ z (1+x)+ (1+x) &=& 24 \\ \mathbf{(1+x)(1+z)} &=& \mathbf{24}\quad (3) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \dfrac{(1)*(3)}{(2)}: & \dfrac{(1+y)(1+x)(1+x)(1+z)} {(1+z)(1+y)} &=& \dfrac{20*24}{30}\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ & (1+x)(1+x)^2 &=& \dfrac{20*24}{30} \\ & (1+x)^2 &=& 16 \\ & 1+x &=& 4 \quad | \quad x > 0 !\\ & \mathbf{x} &=& \mathbf{3} \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline \dfrac{(2)*(1)}{(3)}: & \dfrac{(1+z)(1+y)(1+y)(1+x) } {(1+x)(1+z) } &=& \dfrac{30*20}{24} \\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ & (1+y)(1+y)^2 &=& \dfrac{30*20}{24} \\ & (1+y)^2 &=& 25 \\ & 1+y &=& 5 \quad | \quad y > 0 !\\ & \mathbf{y} &=& \mathbf{4} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \dfrac{(3)*(2)}{(1)}: & \dfrac{(1+x)(1+z)(1+z)(1+y)} {(1+y)(1+x)} &=& \dfrac{24*30}{20} \\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ & (1+z)(1+z)^2 &=& \dfrac{24*30}{20} \\ & (1+z)^2 &=& 36 \\ & 1+z &=& 6 \quad | \quad z > 0 !\\ & \mathbf{z} &=& \mathbf{5} \\ \hline \end{array}\)
