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

##### 2 Answers

#### Explanation:

Let

#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

**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!

#### Explanation:

For starters, you know that

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

#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

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())#