Question #69e1d

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Answer 1 · 2016-10-19T07:21:25

$x = 80/9$

$sqrt(20-x) + 8 = sqrt(9-x) + 11$

$=> sqrt(20-x) = sqrt(9-x)+3$

$=> (sqrt(20-x))^2 = (sqrt(9-x)+3)^2$

$=> 20-x = (sqrt(9-x))^2+2(3)sqrt(9-x)+3^2$

$=> 20-x = 9-x+6sqrt(9-x)+9$

$=> 2 = 6sqrt(9-x)$

$=> sqrt(9-x) = 1/3$

$=> (sqrt(9-x))^2 = (1/3)^2$

$=> 9-x = 1/9$

$=> x = 9-1/9$

$=> x = 80/9$

Checking our result:

$sqrt(20-80/9)+8 = sqrt(100/9)+8$

$=10/3+8$

$=34/3$

$=1/3+11$

$=sqrt(1/9)+11$

$=sqrt(9-80/9) + 11$

as desired.