Given the algebraic relationship $0.02x = 0.5y$. The goal is to find the value of the expression $\frac{x-y}{x+y}$.
Start with the given equation:
$0.02x = 0.5y$To find the ratio $\frac{x}{y}$, divide both sides by $y$ and then by $0.02$:
$\frac{x}{y} = \frac{0.5}{0.02}$Simplify the fraction by multiplying the numerator and denominator by 100:
$\frac{x}{y} = \frac{0.5 \times 100}{0.02 \times 100} = \frac{50}{2} = 25$This shows that $x = 25y$.
Substitute the value of $x$ (in terms of $y$) into the expression $\frac{x-y}{x+y}$:
$\frac{x-y}{x+y} = \frac{(25y)-y}{(25y)+y}$Combine the terms in the numerator and the denominator:
$\frac{25y - y}{25y + y} = \frac{24y}{26y}$Cancel the common factor $y$ from the numerator and denominator (assuming $y \neq 0$):
$\frac{24}{26}$Reduce the fraction to its simplest form by dividing both the numerator and the denominator by their greatest common divisor, which is 2:
$\frac{24 \div 2}{26 \div 2} = \frac{12}{13}$Thus, the value of $\frac{x-y}{x+y}$ is $\frac{12}{13}$.
If \(\frac{x}{y} = \frac{5}{3}\), then \(\frac{x + y}{x - y}\) is equal to
If \(\frac{a}{b} = \frac{7}{6}\), the find the value of the expression \(\frac{6a+13b}{6a-13b}\).
If \(\frac b a = 0.7,\) find the value of \(\frac {a-b}{a+b} + \frac {11}{34}.\)
Consider the following statements:
1. If (a + b) is directly proportional to (a - b), then (a2 + b2) is is directly proportional to ab.
2. If a is directly proportional to b, then (a2 - b2) is directly proportional to ab.
Which of the statements given above is/are correct?