Question

xy/y-x=110 , yz/z-y=132 , zx/z+x=60/11​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

the solutions obtained for x, y, and z after solving equations xy/y-x=110 , yz/z-y=132 , zx/z+x=60/11​ are x=10, y=11, and z=12

Given,

Three equations xy/y-x=110 , yz/z-y=132 , zx/z+x=60/11​

(1) $\frac{xy}{y-x}=110$

(2) $\frac{yz}{z-y}=132$

(3) $\frac{zx}{z+x}=\frac{60}{11}$

To find,

The values of x, y, and z

Solution,

This problem can be solved easily by the following procedure:

First, take the reciprocals of equations (1), (2), and (3)

(1) becomes $\frac{y-x}{xy}=\frac{1}{110}$. Let it be (4)

(2) becomes $\frac{z-y}{yz}=\frac{1}{132}$. Let it be (5)

(3) becomes $\frac{z+x}{zx}=\frac{11}{60}$. Let it be (6)

Now, add (4), (5), and (6) to get equation (7) as follows:

⇒ $\frac{y-x}{xy} + \frac{z-y}{yz} + \frac{z+x}{zx} = \frac{1}{110} + \frac{1}{132} + \frac{11}{60}$

⇒ $\frac{(y-x)*z + (z-y)*x + (z+x)*y}{xyz} = \frac{6+5+11*11}{660}$     [by taking LCM and multiplying on both sides]

⇒ $\frac{yz-xz+xz-xy+yz+xy}{xyz} = \frac{6+5+121}{660}$  

⇒ $\frac{2yz}{xyz} = \frac{132}{660}$  

⇒ $\frac{2}{x} = \frac{1}{5}$

⇒ $\frac{x}{2} = 5$

⇒ $x = 5*2$

⇒ $x=10$

Therefore, we have found the value of x=10

Now, substitute the value of x in equation (1) to obtain the value of y

⇒ $\frac{10y}{y-10}=110$

⇒ $10y=110(y-10)$

⇒ $10y=110y-1100$

⇒ $110y-10y=1100$

⇒ $100y=1100$

⇒ $y=\frac{1100}{100}$

⇒ $y=11$

Therefore, we have found the value of y=11

Now, substitute the value of y in equation (2) to obtain the value of z

⇒ $\frac{11z}{z-11}=132$

⇒ $11z=132(z-11)$

⇒ $11z=132z-1452$

⇒ $132z-11z=1452$

⇒ $121z=1452$

⇒ $z=\frac{1452}{121}$

⇒ $z=1$

Therefore, we have found the value of z=12

Hence, the solutions obtained for x, y, and z after solving equations (1), (2), and (3) are x=10, y=11, and z=12