The value that is a solution of the rational equation that must be discarded because of the context is (b) x = -10
The equation is given as:
[tex]\frac 1x + \frac{1}{x + 18} = \frac{1}{40}[/tex]
Take the LCM
[tex]\frac{x + 18 + x}{x(x + 18)} = \frac{1}{40}[/tex]
This gives
[tex]\frac{2x + 18}{x(x + 18)} = \frac{1}{40}[/tex]
Cross multiply
[tex]40(2x + 18) =x(x + 18)[/tex]
Expand
[tex]80x + 720 =x^2 + 18x[/tex]
Collect like terms
[tex]x^2 + 18x - 80x -720 = 0[/tex]
[tex]x^2 - 62x -720 = 0[/tex]
Using a graphing calculator, we have:
[tex]x = 72\ or\ x =-10[/tex]
The value of x cannot be negative, because of the context.
So, the value that is a solution of the rational equation that must be discarded because of the context is (b) x = -10
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