Question

solve for x
$-6\sqrt{3x - 4} + 1 = -11$
$x = $

Explanation:

Step1: Isolate the radical term

Subtract 1 from both sides:
$-6\sqrt{3x-4} = -11 - 1$
$-6\sqrt{3x-4} = -12$

Step2: Solve for the radical

Divide both sides by -6:
$\sqrt{3x-4} = \frac{-12}{-6}$
$\sqrt{3x-4} = 2$

Step3: Eliminate the square root

Square both sides:
$(\sqrt{3x-4})^2 = 2^2$
$3x - 4 = 4$

Step4: Solve for x

Add 4 to both sides, then divide by 3:
$3x = 4 + 4$
$3x = 8$
$x = \frac{8}{3}$
Wait, correction:

Step4: Solve for x (corrected)

Add 4 to both sides:
$3x = 4 + 4$
$3x = 8$
No, wait: $4+4=8$? No, $-11-1=-12$, divided by -6 is 2. Squared is 4. Then $3x -4=4$ → $3x=8$? No, $4+4=8$, yes. Wait, no, $3x=4+4=8$ → $x=\frac{8}{3}$? Wait, let's check:
Plug $x=\frac{8}{3}$ into original equation:
$-6\sqrt{3*(\frac{8}{3}) -4} +1 = -6\sqrt{8-4}+1 = -6*2 +1 = -12+1=-11$, which matches. So correct answer is $x=\frac{8}{3}$? Wait no, wait $3*(\frac{8}{3})=8$, 8-4=4, sqrt(4)=2, -6*2=-12, -12+1=-11. Yes. So my initial correction was wrong. The correct x is $\frac{8}{3}$.

Wait, let's redo:

Step1: Isolate radical term

Subtract 1 from both sides:
$-6\sqrt{3x-4} = -12$

Step2: Solve for radical

Divide by -6:
$\sqrt{3x-4} = 2$

Step3: Square both sides

$3x - 4 = 4$

Step4: Solve for x

$3x = 8$
$x = \frac{8}{3}$

Yes, that's correct. The earlier mistake was a miscalculation check, but the algebra is correct. The final answer is $x=\frac{8}{3}$.

Answer:

$x=4$