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.

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

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erinna erinna · Answer 2 · 2019-09-24T14:01:05+02:00

Answer:

[tex]y=3x+\frac{2}{x}[/tex]

Step-by-step explanation:

If y is directly proportional to x, then

[tex]y\propto x[/tex]

[tex]y=ax[/tex]

where, a is constant of proportionality.

If y is inversely proportional to x, then

[tex]y\propto \frac{1}{x}[/tex]

[tex]y=\frac{b}{x}[/tex]

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.

[tex]y=ax-\frac{b}{x}[/tex] ..... (1)

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

[tex]-7=a(-2)-\frac{b}{-2}[/tex]

[tex]-7=-2a+\frac{b}{2}[/tex]

[tex]-7=\frac{-4a+b}{2}[/tex]

Multiply both sides by 2.

[tex]-14=-4a+b[/tex] .... (2)

We have y=5 at x=1.

[tex]5=a(1)-\frac{b}{1}[/tex]

[tex]5=a-b[/tex] .... (3)

Add equation (2) and (3).

[tex]-14+5=-4a+b+a-b[/tex]

[tex]-9=-3a[/tex]

Divide both sides by -3.

[tex]3=a[/tex]

The value of a is 3.

Substitute a=3 in equation (3).

[tex]5=3-b[/tex]

[tex]5-3=-b[/tex]

[tex]2=-b[/tex]

[tex]-2=b[/tex]

The value of b is -2.

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

[tex]y=(3)x-\frac{(-2)}{x}[/tex]

[tex]y=3x+\frac{2}{x}[/tex]

Therefore the required equation is [tex]y=3x+\frac{2}{x}[/tex].