How do you solve $\sqrt{25} - 3 x - 6 = 1$ $sqrt25 - 3x -6 = 1$ ?

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Answer 1 · 2015-07-24T18:29:30

You isolate $x$ $x$ on one side of the equation.

The thing to notice about this equation is that $\sqrt{25}$ $sqrt(25)$ is actually equal to

$\sqrt{25} = \sqrt{5^{2}} = 5$ $sqrt(25) = sqrt(5""^2) = 5$

This means that your original equation becomes

$5 - 3 x - 6 = 1$ $5 - 3x - 6 = 1$

You can easily solve this by first isolating $- 3 x$ $-3x$ on one side of the equation first. Start from

$- 1 - 3 x = 1$ $-1 - 3x = 1$

Add $+ 1$ $+1$ to both sides of the equation to get

$cancel(-1) + cancel(1) - 3x = 1 + 1$

$-3x = 2$

Now divide both sides of the equation by $-3$ to get

$(cancel(-3) * x)/cancel(-3) = 2/-3$

$x = color(green)(-2/3)$

How do you solve sqrt25 - 3x -6 = 1√25−3x−6=1sqrt25 - 3x -6 = 1?