The break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system (x) = 500,000 + 400x |R(x) = 600x [y = 500,000 + 400x or ly = 600x. Using the substitution method, we can substitute 600x for y in the first equation. 600x = 500,000 + 400x Substitute 600x for y in y= 500,000 + 400x. 200x 500,000 Subtract 400x from both sides. x = 2500 Divide both sides by 200. Back-substituting 2500 for x in either of the system's equations (or functions), we obtain R(2500) = 600(2500) = 1,500,000. We used R[x) = 60ox.

The break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system (x) = 500,000 + 400x |R(x) = 600x [y = 500,000 + 400x or ly = 600x. Using the substitution method, we can substitute 600x for y in the first equation. 600x = 500,000 + 400x Substitute 600x for y in y= 500,000 + 400x. 200x 500,000 Subtract 400x from both sides. x = 2500 Divide both sides by 200. Back-substituting 2500 for x in either of the system's equations (or functions), we obtain R(2500) = 600(2500) = 1,500,000. We used R[x) = 60ox.

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

The break-even point occurs where the graphs of C and R intersect. Thus, we find
this point by solving the system
(x) = 500,000 + 400x
|R(x) = 600x
[y = 500,000 + 400x
or
ly = 600x.
Using the substitution method, we can substitute 600x for y in the first equation.
600x = 500,000 + 400x Substitute 600x for y in y= 500,000 + 400x.
200x
500,000
Subtract 400x from both sides.
x = 2500
Divide both sides by 200.
Back-substituting 2500 for x in either of the system's equations (or functions), we
obtain
R(2500) = 600(2500) = 1,500,000.
We used R[x) = 60ox.

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