How do you solve this? √3x+1 - √2x-1 = 1?

√3x+1 - √2x-1 = 1
√3x+1=1-√(2x-1)
(√3x+1)^2=(1-√2x-1)^2
3x+1=1-2√(2x-1)+2x-1
3x+1=-2√(2x-1)+2x
3x-2x+1=-2√(2x-1)
x+1=-2√(2x-1)
(x+1)^2=(-2√(2x-1))^2
x^2+2x+1=4(2x-1)
x^2+2x+1=8x-4
x^2+2x-8x+1+4=0
x^2-6x+5=0
(x-1)(x-5)=0
x-1=0 x-5=0
x=1 x=5 answer roots are {1,5} @(^_^)@

√(3x + 1) - √(2x - 1) = 1
√(3x + 1) = 1 + √(2x - 1)
(√(3x + 1))² = (1 + √(2x - 1))²
3x + 1 = (1 + √(2x - 1))(1 + √(2x - 1))
3x + 1 = 1 + 2√(2x - 1) + 2x - 1
3x + 1 = 2√(2x - 1) + 2x
x + 1 = 2√(2x - 1)

(x + 1)² = (2√(2x - 1))²
x² + 2x + 1 = 4(2x - 1)
x² + 2x + 1 = 8x - 4
x² - 6x + 5 = 0
x² - x - 5x + 5 = 0
x(x - 1) - 5(x - 1) = 0
(x - 1)(x - 5) = 0

x - 1 = 0 → x = 1
x - 5 = 0 → x = 5


Verify the answer:
√(3x + 1) - √(2x - 1) = 1
√(3*1 + 1) - √(2*1 - 1) = 1
√(3 + 1) - √(2 - 1) = 1
√(4) - √(1) = 1
2 - 1 = 1
1 = 1

√(3x + 1) - √(2x - 1) = 1
√(3*5 + 1) - √(2*5 - 1) = 1
√(15 + 1) - √(10 - 1) = 1
√(16) - √(9) = 1
4 - 3 = 1
1 = 1

√(3x + 1) - √(2x - 1) = 1
√(3x + 1) = 1 + √(2x - 1)
3x + 1 = [1 + √(2x - 1)]^2
3x + 1 = [1 + √(2x - 1)][1 + √(2x - 1)]
3x + 1 = 1 + 1√(2x - 1) + 1√(2x - 1) + (2x - 1)
3x + 1 = 1 + 2√(2x - 1) + 2x - 1
2√(2x - 1) = 3x + 1 - 1 - 2x + 1
2√(2x - 1) = x + 1
√(2x - 1) = (x + 1)/2
2x - 1 = [(x + 1)/2]^2
2x - 1 = (x + 1)(x + 1)/4
4(2x - 1) = x*x + 1*x + x*1 + 1*1
8x - 4 = x^2 + x + x + 1
x^2 + 2x - 8x + 1 + 4 = 0
x^2 - 6x + 5 = 0
x^2 - x - 5x + 5 = 0
(x^2 - x) - (5x - 5) = 0
x(x - 1) - 5(x - 1) = 0
(x - 1)(x - 5) = 0

x - 1 = 0
x = 1

x - 5 = 0
x = 5

∴ x = 1 , 5

√(3x+1) - √(2x-1) = 1
Squaring both sides
3x + 1 - 2√(3x+1)(2x-1) + 2x - 1 = 1
5x - 1 = 2√(3x+1)(2x-1)
Squaring again
25x^2 - 10x + 1 = 4(6x^2 - x - 1)
25x^2 - 10x + 1 = 24x^2 - 4x - 4
x^2 - 6x + 5 = 0
(x - 1)(x - 5) = 0
(x - 1) = 0
x = 1
(x - 5) = 0
x = 5
x = {1, 5}

√3x+1 - √2x-1 = 1

then

(√3x+1 - √2x-1 )² = 1²
(3x + 1) - 2√(3x+1)(2x-1) + (2x - 1) = 1
5x - 1 = 2√(3x+1)(2x-1)

then

(5x - 1)² = (2√(3x+1)(2x-1))²
25x² - 10x + 1 = 4(3x+1)(2x - 1)
25x² - 10x + 1 = 24x² - 4x - 4
x² - 6x + 5 = 0
(x - 5)(x - 1) = 0

then

x = 5 or x = 1

([x + a million]/[3x - a million]) + ([2x + a million]/[3x - 2]) = - a million ([x + a million][3x - 2]) + ([2x + a million][3x - a million]) = - ([3x - a million][3x - 2]) 3x² - 2x + 3x - 2 + 6x² - 2x + 3x - a million = - (9x² - 6x - 3x + 2) 9x² + 2x - 3 = - 9x² + 9x - 2 18x² - 7x = a million x² - 7/36x = a million/18 + (- 7/36)² x² - 7/36x = seventy two/a million,296 + 40 9/a million,296 (x - 7/36)² = 121/a million,296 x - 7/36 = +/- 11/36 x = 11/36 + 7/36, x = 18/36, x = a million/2 x = - 11/36 + 7/36, x = - 4/36, x = - a million/9 answer: x = a million/2, - a million/9 evidence (x = a million/2 or 0.5): ([0.5 + a million]/[3{0.5} - a million]) + ([2{0.5} + a million]/[3{0.5} - 2]) = - a million (a million.5/[a million.5 - a million]) + ([a million + a million]/[a million.5 - 2]) = - a million (a million.5/0.5) + (2/- 0.5) = - a million 3 - 4 = - a million you'll be able to coach x = - a million/9 an identical way.

sqrt.3x+1 - sqrt.2x-1 = 1
square both sides
(3x+1) - (2x-1) =1
multiply the - through 2x-1
3x+1 -2x+1=1
simplify
x+2=1
subtract 2 to other side
x=-1

It seems that "Answers" does not admit the use of the radical sign, so rather than show you how to solve the problem I will tell you step by step how to do it

1). Move the sq rt(2x - 1) to the right side of the equation.

2) Square both sides of the equation.

3) Collect terms leaving the (2)sq rt(2x - 1) on the right side of the equation.

4). Square both sides of the equation again and collect like terms on the left side giving x^2 - 6x + 5 = 0.

5). Factor this last quadratic using the quadratic solution formula or any other method to find that the quadratic in 4) factors as
((x - 1)(x - 5), from which the solutions are x = 1 and x = 5. You should verify this by plugging these values into the original equation.

√3x+1 - √2x-1 = 1
√3x+1 = √2x-1 + 1
square both sides, simplify then square both sides again ...etc

x = 1,5