Question

Let y be the difference of two numbers such that one number varies directly as x while the other number varies inversely as x. If y = –7 when x = –2, and y = 5 when x = 1, express y in terms of x. Show your work.

Answer 1

Let a and b the two numbers. So y = a – b. one number varies directly as x , a= k1x. while the other number varies inversely as x, b = k2/x. so the equation y = k1x – k2/x
At y = -7, x =2 then -7 = -2k1 + k2/2
At  y = 5, x =1 then 5 = k1 – k2
Solving k1 and k2, 3 and -2 respectively
So y = 3x+2/x

Answer 2

Answer:

$y=3x+\frac{2}{x}$

Step-by-step explanation:

If y is directly proportional to x, then

$y\propto x$

$y=ax$

where, a is constant of proportionality.

If y is inversely proportional to x, then

$y\propto \frac{1}{x}$

$y=\frac{b}{x}$

where, b is constant of proportionality.

It is given that y be the difference of two numbers such that one number varies directly as x while the other number varies inversely as x.

$y=ax-\frac{b}{x}$            ..... (1)

We have y=-7 at x=-2.

$-7=a(-2)-\frac{b}{-2}$

$-7=-2a+\frac{b}{2}$

$-7=\frac{-4a+b}{2}$

Multiply both sides by 2.

$-14=-4a+b$                  .... (2)

We have y=5 at x=1.

$5=a(1)-\frac{b}{1}$

$5=a-b$                 .... (3)

Add equation (2) and (3).

$-14+5=-4a+b+a-b$

$-9=-3a$

Divide both sides by -3.

$3=a$

The value of a is 3.

Substitute a=3 in equation (3).

$5=3-b$

$5-3=-b$

$2=-b$

$-2=b$

The value of b is -2.

Substitute a=3 and b=-2 in equation (1).

$y=(3)x-\frac{(-2)}{x}$

$y=3x+\frac{2}{x}$

Therefore the required equation is $y=3x+\frac{2}{x}$.