how to you find 7 times the square root of 3x-4 (3x-4 is under the square root sign) + 7 =35

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Let sqrt = square root for short. 7(sqrt{3x - 4}) + 7 = 35 This is a radical equation. The goal is to isolate the radical on one side of the equation. Subtract 7 from both sides as step one. Doing so, we get this: 7(sqrt{3x - 4} = 28 We now divide both sides by 7. This will isolate the radical on the left side. sqrt{3x - 4} = 28/7 sqrt{3x - 4} = 4 We now square both sides. This is remove the radical from the left side of the equation. (sqrt{3x - 4})^2 = (4)^2 3x - 4 = 16 3x = 16 + 4 3x = 20 x = 20/3 Is this the true value of x? We must prove this is the right value of x by plugging 20/3 for x in the original question given. Afterward, we simplify the radical equation. The idea is get the same answer on both sides. Let x = 20/3 7(sqrt{3(20/3) - 4}) + 7 = 35 7(sqrt{20 - 4}) + 7 = 35 7(sqrt{16}) + 7 = 35 7(4) + 7 = 35 28 + 7 = 35 35 = 35...It checks!!! So, the right answer for x is 20/3.