Question

If `2((3x+ 1)/(2x- 5))^2 + (3x + 1)/(2x- 5) - 6= 0`, then what is the value of `x`?

Answer 1

`x = 9/7`

Explanation:

Let `u = (3x + 1)/(2x - 5)`. Then the equation becomes:

`2u^2 + u - 6 = 0`

`2u^2 + 4u - 3u - 6 = 0`

`2u(u + 2) - 3(u + 2) = 0`

`(2u - 3)(u + 2) =0`

`u = 3/2 or -2`

We now return to our original variable `x`.

Case 1: `u = 3/2`

`3/2 = (3x + 1)/(2x - 5)`

`3(2x - 5) = 2(3x + 1)`

`6x - 15 = 6x + 2`

`0x = 17`

`x = O/`

Case 2: `u = -2`

`-2 = (3x + 1)/(2x - 5)`

`-2(2x - 5) = 3x + 1`

`-4x + 10 = 3x + 1`

`9 = 7x`

`x = 9/7`

Hopefully this helps!

Answer 2

`x = 9/7`

Explanation:

For starters, you know that `x` cannot be equal to `5/2` because that would make the two denominators equal to zero.

In other words, you need to have

`2x - 5 != 0`

`x != 5/2`

Now, let's say that

`(3x+ 1)/(2x - 5) = y" "color(darkorange)("(*)")`

The original equation becomes

`2y^2 + y - 6 = 0`

You can solve this quadratic equation by using the quadratic formula, which for a general-form quadratic equation

`color(blue)(ul(color(black)(ax^2 + bx + c = 0)))`

looks like this

`color(blue)(ul(color(black)(x_(1,2) = (-b +- sqrt(b^2 - 4 * a * c))/(2 * a))))`

In your case, you have

`{(a = 2), (b = 1), (c = - 6) :}`

and so

`y_(1,2) = (-1 +- sqrt( 1^2 - 4 * 2 * (-6)))/(2 * 2)`

`y_(1,2) = (-1 +- sqrt(1 + 48))/4`

`y_(1,2) = (-1 +- sqrt(49))/4 implies {( y_1 = (-1 - 7)/4 = -2), (y_2 = (-1 + 7)/4 = 3/2) :}`

Take both values of `y` and substitute them back into `color(darkorange)("(*)")` to get

• `ul("For" color(white)(.)y = -2)`

`(3x + 1)/(2x - 5) = -2`

This is equivalent to

`3x + 1 = -2 * (2x - 5)`

`3x + 1 = - 4x + 10`

which gets you

`7x = 9 implies x = 9/7`

• `ul("For" color(white)(.)y = 3/2)`

`(3x + 1)/(2x - 5) = 3/2`

This is equivalent to

`3x + 1 = 3/2 * (2x - 5)`

`color(red)(cancel(color(black)(3x))) + 1 = color(red)(cancel(color(black)(3x))) - 15/2`

`1 != -15/2`

Therefore, you can say that the original equation has `1` valid solution, `x = 9/7`. Notice that this solution satisfies the initial condition.

Do a quick double-check to make sure that the calculations are correct

`2 * ((3 * 9/7 + 1)/(2 * 9/7 - 5))^2 + (3 * 9/7 + 1)/(2 * 9/7 - 5) - 6 = 0`

`2 * (((27 + 7)/color(red)(cancel(color(black)(7))) )/( (18 - 35)/color(red)(cancel(color(black)(7)))))^2 + ((27 + 7)/color(red)(cancel(color(black)(7))) )/( (18 - 35)/color(red)(cancel(color(black)(7)))) - 6 = 0`

`2 * (34/(-17))^2 + (34/(-17)) - 6 = 0`

`2 * (-2)^2 + (- 2) - 6 = 0`

`2 * 4 - 8 = 0 " " color(darkgreen)(sqrt())`