How do you solve $2sqrt(3x+5)+7=16$ ?

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Answer 1 · 2017-05-09T02:27:55

See a solution process below:

First, subtract $color(red)(7)$ from each side of the equation to isolate the radical term while keeping the equation balanced:

$2sqrt(3x + 5) + 7 - color(red)(7) = 16 - color(red)(7)$

$2sqrt(3x + 5) + 0 = 9$

$2sqrt(3x + 5) = 9$

Next, divide each side of the equation by $color(red)(2)$ to isolate the radical while keeping the equation balanced:

$(2sqrt(3x + 5))/color(red)(2) = 9/color(red)(2)$

$(color(red)(cancel(color(black)(2)))sqrt(3x + 5))/cancel(color(red)(2)) = 9/2$

$sqrt(3x + 5) = 9/2$

Then, square each side of the equation to eliminate the radical while keeping the equation balanced:

$(sqrt(3x + 5))^2 = (9/2)^2$

$3x + 5 = 81/4$

Next, subtract $color(red)(5)$ from each side of the equation to isolate the $x$ term:

$3x + 5 - color(red)(5) = 81/4 - color(red)(5)$

$3x + 0 = 81/4 - (4/4 xx color(red)(5))$

$3x = 81/4 - 20/4$

$3x = 61/4$

$now, divide each side of the equation by$ color(red)(3) $to solve for$ x# while keeping the equation balanced:

$(3x)/color(red)(3) = (61/4)/color(red)(3)$

$(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 61/12$

$x = 61/12$

How do you solve 2sqrt(3x+5)+7=162sqrt(3x+5)+7=16?