Question

Help i dont understand how to complete this

Answer 1

Answer: Josh = 19.8 hours, Danny = 10.8 hours

Explanation:

`Josh: (1)/(x+9)\\\\\\Danny: (1)/(x)\\\\\\Together: (1)/(7)\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\(1)/(x+9)\quad +\qquad (1)/(x)\qquad = \qquad (1)/(7)\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=(-b\pm √(b^2-4ac))/(2a)`

`x=(-(-5)\pm √((-5)^2-4(1)(-63)))/(2(1))\\\\\\.\quad =(5\pm√(25+252))/(2)\\\\\\.\quad =(5\pm√(277))/(2)\\\\\\.\quad =(5\pm 16.6)/(2)\\\\\\x =(5+16.6)/(2)\qquad x=(5-16.6)/(2)\\\\\\x=(21.6)/(2)\qquad \qquad x=(-11.6)/(2)\\\\\\x=10.8 \qquad \qquad x=-5.8`

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9 --> 10.8 + 9 = 19.8

Danny: x --> 10.8

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2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 (x = -1 is an extraneous solution so is not valid)

2d) No Solution (x = 1 is an extraneous solution so is not valid)

Here is the work for 2a. Follow this format for b, c, & d

`(3)/(k^2-8x+12)=(k)/(k-2)-(4)/(k-6)\\\\\text{Since the denominator cannot equal zero, then } k \\eq2\ and\ k\\eq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}`

`x=(-(-10)\pm √((-10)^2-4(1)(5)))/(2(1))\\\\\\.\quad =(10\pm√(100-20))/(2)\\\\\\.\quad =(10\pm√(80))/(2)\\\\\\.\quad =(10\pm8.9)/(2)\\\\\\x=(10+8.9)/(2)\qquad x=(10-8.9)/(2)\\\\\\x=(18.9)/(2)\qquad x=(1.1)/(2)\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}`