Question

a. The sum of the product and quotient of two numbers is equal to four times their difference.


b. The age of Kevin now is two times the age of his younger brother three years ago.


c. Subtracting two from both the numerator and denominator of an unknown fraction yields to an answer of 1

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Answer 1

Let Kevin's age be x

Let his brother's age be x - 3

Age of kevin

2 ( x - 3 ) = 2x - 6

c )

$\frac{x - 1}{y - 1} = 1$

x = 2

y = 2

Answer 2

a. Mathematical expression is $\boxed{xy+\dfrac{x}{y}=4(x-y)}$

b. Mathematical expression is $\boxed{x=2(y-3)}$

c. Mathematical expression is $\boxed{\dfrac{x-2}{y-2}=1}$

Step-by-step explanation:

a. The sum of the product and quotient of two numbers is equal to four times their difference.

Solution:

Let the two numbers be $x$ and $y$ with $x>y$.

Then their product is $xy$, quotient is $\dfrac{x}{y}$ and difference is $(x-y)$.

By the given conditions,

$\quad xy+\dfrac{x}{y}=4(x-y)$

This is the required mathematical expression.

b. The age of Kevin now is two times the age of his younger brother three years ago.

Solution:

Let, Kevin is $x$ years old now and Kevin's younger brother is $y$ years old now.

Then three years ago, Kevin's younger brother was $(y-3)$ years old.

According to the condition,

$\quad x=2\times (y-3)$

$\Rightarrow x=2(y-3)$

This is the required mathematical expression.

c. Subtracting two from both the numerator and denominator of an unknown fraction yields to an answer of 1.

Solution:

Let the fraction be $\dfrac{x}{y}$ where $x$, $y$ are integers and $y\neq 0$.

When $2$ is subtracted from both the numerator and the denominator, the new fraction becomes $\dfrac{x-2}{y-2}$.

According to the question,

$\quad \dfrac{x-2}{y-2}=1$

This is the required mathematical expression.