can anyone help with solving the following linear equation?

t = 240/r

r = t + 22

Replace t with 240/r in the second equation

r = 240/r + 22

Multiply both sides by r

r^2 = 240 + 22r

Subtract 22r from each side, and subtract 240 from each side

r^2 - 22r - 240 = 0

Factor

(r-30)(r+8) = 0

Set each factor equal to zero

r-30 = 0 or r+8 = 0

r = 30 or r = -8

Now find the corresponding t value.

t = 240/r

So t = 240/30 = 8 or t = 240/-8 = -30

Answer: r = 30, t = 8,

or r = -8, t = -30

Substitute one equation into the other.

1. t=240/r

2. r=t+22

Rearrange equation 2 to give

t= r-22 and substitute into 1.

r-22=240/r

Multiply by r each side

r^2 - 22r = 240

or

r^2 -22r -240 = 0

Using r = (-b +/- sqrt (b^2-4ac) )/ 2a

r= ( 22 +/- sqrt(22^2 -4*1*(-240) )/ 2(1)

r= (22 +/- sqrt (484+960))/2

= (22+/- sqrt(1444))/2 = (22 +/- 38)/2

so r = (22+38)/2 = 30 or r= (22-38)/2 = -8

so for r = 30, t= 8 and for r= -8, t = -30

Two solutions

Verify in each equation to ensure that this is correct!

You can solve it by graphing the two lines and detect the point of intersection

or by algebra

t = 240/r equ. 1

r = t+22 equ. 2

by substitute from equ. 1 in equ. 2

r =(240/r) + 22

multiply all the equ. By (r)

r^2 - 22r - 240 = 0

factors

(r+8)(r-30) = 0

then we have

r=-80 OR r=30

i hope you get it

this is my best to explain

thanx bye