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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 for College Students
Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter13: Conic Sections
Section13.5: Systems Involving Nonlinear Equations
Problem 39PS
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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.
Transcribed Image Text: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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