anemone said:Let $x,\,y,\,z$ be real numbers which satisfy the system below:
$x+\dfrac{1}{yz}=\dfrac{1}{5}$
$y+\dfrac{1}{xz}=-\dfrac{1}{15}$
$z+\dfrac{1}{xy}=\dfrac{1}{3}$
Evaluate $\dfrac{z-y}{z-x}$.
kaliprasad said:clearly x,y,z none of them is zero deviding 1st equation by x 2nd by y and 3rd by z we get
$1+\dfrac{1}{xyz}= \dfrac{1}{5x}=\dfrac{1}{15y} = \dfrac{1}{3z}$
hence $5x=15y=3z$
or $\dfrac{x}{z}= \dfrac{3}{5}$
$\dfrac{y}{z}= \dfrac{1}{5}$
hence $\dfrac{z-y}{z-x}=\dfrac{1-\frac{y}{z}}{1-\frac{x}{z}}=\dfrac{1-\frac{1}{5}}{1-\frac{3}{5}}=2 $
anemone said:Thanks kaliprasad for participating...but:
Please note that the RHS of the second equation has a minus sign..
In this context, "evaluate" means to find the numerical value of the expression (z-y)/(z-x) given specific values for the variables x, y, and z.
Yes, the expression can be simplified by factoring out a common factor of (z-x) in the numerator and denominator, resulting in (z-y)/(z-x) = 1.
Yes, the expression can have a real number as its value as long as the denominator (z-x) is not equal to 0. In that case, the value would be undefined.
No, there are no specific restrictions on the values of x, y, and z. They can be any real numbers.
The expression (z-y)/(z-x) can be used in various real-life situations, such as calculating the slope between two points on a graph, determining the rate of change in a mathematical model, or finding the average rate of change between two data points in a real-world scenario.