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Solve each equation. Check your solutions.$$\frac{4}{3-p}+\frac{2}{5-p}=\frac{26}{15}$$
Precalculus
Algebra
Chapter 11
Quadratic Equations, Inequalities, and Functions
Section 4
Equations Quadratic in Form
Introduction to Conic Sections
Equations and Inequalities
Functions
Polynomials
Harvey Mudd College
University of Michigan - Ann Arbor
Lectures
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for exercise 16. We're going to multiply by our denominators. Denominators. So we're going to have 300 finest 60 P plus 4 50 minus 30 P. What did I say? 450. Mhm. That should be 90. Mhm. And Then that'll be -30p. Equal to. And I'm just going to write it this way 15 minus eight P. Plus P squared. And so when we multiply 26 It's going to be 26 p. Squared minus eight times 26 Is -208 p. 15 times 26. There's 3 90. Then our left hand side when we combine like terms we have negative 90 p Plus 390. We need everything to the right hand side. We have 26 p squared -1 18 p. Then we can write this as mhm. Let's see it's two P. 13 P minus 59. Then we can say Pecan equal in this situation zero or 59 13th. Now let's check. So if P equals zero that's going to be For this exercise to over one Plus 3/2 equals this. Sorry I meant 4/3 plus 2/5 Equal to 26/15. All right so that means R. L. c. d. here is going to be 15. So it's going to be 20-plus 6 which is 26. So that's true. Now let's try plugging in 59 13. So we have four over 3 -59/13 plus two over 5 -59/13. Equal to 26/15. So in other words we have four over negative 33rd plus two over 6, 13 Equal to 26, 15. We can rewrite this as 52 over negative 20 Plus 26/6. You go to 26/15 and 30 is going to be our L. c. d. here. So we're going to have negative 78 plus 130. I think that's going to equal 52 over 30 Is equal to 26/15. Which is true. Therefore, solutions are true.
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