math stuck on a 2 different question answer asap please and thank you!

Question

√6x  + 7 =√7x+7   +6                  Note only 6x is under radical on left side and  7x+7 are both under radical on right side

 

and..

 

 

7√x²+3x-1  +8=5                         Note X²+3X-1   Are all under the radical,   +8 is not.

Answer 1

For the first one.

 

√(6x) + 7 = √(7x + 7) + 6

 

 

Subtract 7 on both sides of equation.

 

√(6x) = √(7x + 7) - 1

 

 

Square both sides of equation.

 

6x = (√(7x + 7) - 1)²

 

6x = 7x + 7 - 2√(7x + 7) + 1

 

6x = 7x + 8 - 2√(7x + 7)

 

 

Subtract 7x  and subtract 8 on both sides of equation.

 

-x - 8 = - 2√(7x + 7)

 

 

Divide both sides of equation by -2.

 

(1/2)x + 4 = √(7x + 7)

 

 

Square both sides of equation.

 

((1/2)x + 4)² = 7x + 7

 

(1/4)x² + 4x + 16 = 7x + 7

 

 

Subtract 7x and subtract 7 on both sides of equation.

 

(1/4)x² - 3x + 9 = 0

 

 

Multiply both sides of equation by 4 to get rid of any denominators.

 

x² - 12x - 36 = 0

 

(x - 6)(x - 6) = 0

 

x = 6

 

 

Second problem

 

7√(x² + 3x - 1) + 8 = 5

 

 

Subtract 8 on both sides of equation.

 

7√(x² + 3x - 1) = -3

 

 

√(x² + 3x - 1) = -3/7

 

 

Square both sides of equation.

 

x² + 3x - 1 = 9 / 49

 

 

Multiply both sides of equation by 49.

 

49x² + 147x - 49 = 9

 

 

Subtract 9 on both sides of equation.

 

49x² + 147x - 58 = 0

 

 

Use the quadratic formula to solve for x.  I will let you take it from here.