How do you solve $2root3(7b-1)-4=0$ ?

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Answer 1 · 2017-06-27T13:09:35

See a solution process below:

First, add $color(red)(4)$ to each side of the equation to isolate the radical term while keeping the equation balanced:

$2root(3)(7b - 1) - 4 + color(red)(4) = 0 + color(red)(4)$

$2root(3)(7b - 1) - 0 = 4$

$2root(3)(7b - 1) = 4$

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

$(2root(3)(7b - 1))/color(red)(2) = 4/color(red)(2)$

$(color(red)(cancel(color(black)(2)))root(3)(7b - 1))/cancel(color(red)(2)) = 2$

$root(3)(7b - 1) = 2$

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

$(root(3)(7b - 1))^color(red)(3) = 2^color(red)(3)$

$7b - 1 = 8$

Next, add $color(red)(1)$ to each side of the equation to isolate the $b$ term while keeping the equation balanced:

$7b - 1 + color(red)(1) = 8 + color(red)(1)$

$7b - 0 = 9$

$7b = 9$

Now, divide each side of the equation by $color(red)(7)$ to solve for $b$ while keeping the equation balanced:

$(7b)/color(red)(7) = 9/color(red)(7)$

$(color(red)(cancel(color(black)(7)))b)/cancel(color(red)(7)) = 9/7$

$b = 9/7$

How do you solve 2root3(7b-1)-4=02root3(7b-1)-4=0?