How do you solve #6/x + 7/(x + 3) = 4#?

The first step to answering this is multiplying the whole equation by #x#:

#6 + (7x) / (x+3) = 4x #

Then multiply the whole equation by #x+3# :

# 6(x+3) + 7x = 4x(x+3) #

Expanding;

#6x + 18 + 7x = 4x^2 + 12x #

Rearanging:

#4x^2 - x - 18 = 0 #

Now we have a standard quadratic equation, applying the quadratic formula or factorising to yeild:

#x= 9/4 , x = -2 #

First, multiply each side of the equation by #color(red)(x)(color(blue)(x + 3))# to eliminate the fractions while keeping the equation balanced:

#color(red)(x)(color(blue)(x + 3))(6/x + 7/(x + 3)) = color(red)(x)(color(blue)(x + 3))4#

#(color(red)(x)(color(blue)(x + 3)) xx 6/x) + (color(red)(x)(color(blue)(x + 3)) xx 7/(x + 3)) = (x^2 + 3x)4#

#(cancel(color(red)(x))(color(blue)(x + 3)) xx 6/color(red)(cancel(color(black)(x)))) + (color(red)(x)cancel((color(blue)(x + 3))) xx 7/color(blue)(cancel(color(black)(x + 3)))) = 4x^2 + 12x#

#6(color(blue)(x + 3)) + 7color(red)(x) = 4x^2 + 12x#

#6x + 18 + 7x = 4x^2 + 12x#

#6x + 7x + 18 = 4x^2 + 12x#

#13x + 18 = 4x^2 + 12x#

We can now put the equation in standard form:

#13x - color(red)(13x) + 18 - color(blue)(18) = 4x^2 + 12x - color(red)(13x) - color(blue)(18)#

#0 + 0 = 4x^2 + (12 - color(red)(13))x - 18#

#0 = 4x^2 + (-1)x - 18#

#0 = 4x^2 - 1x - 18#

#4x^2 - 1x - 18 = 0#

We can now factor the left side of the equation as:

#(4x - 9)(x + 2) = 0#

Now, solve each term on the left for #0# to find the solutions to the problem:

Solution 1:

#4x - 9 = 0#

#4x - 9 + color(red)(9) = 0 + color(red)(9)#

#4x - 0 = 9#

#4x = 9#

#(4x)/color(red)(4) = 9/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 9/4#

#x = 9/4#

Solution 2:

#x + 2 = 0#

#x + 2 - color(red)(2) = 0 - color(red)(2)#

#x + 0 = -2#

#x = -2#

The Solution Are: #x = {-2, 9/4}#

#6/x + 7/(x+3) = 4#

The first thing you want to do is to make the denominators the same. How to do that? Well, let's multiply by one, which shouldn't do anything wrong, right?:

#6/x * 1 + 7/(x+3) * 1 = 4#

Now, any number divided by itself equals one, doesn't it? Use this to your advantage! Let's change the one on the left into #(x + 3)/(x + 3)#:

#6/x * (x + 3)/(x + 3) + 7/(x+3) * 1 = 4#

And the one on the right into #x/x#:

#6/x * (x + 3)/(x + 3) + 7/(x+3) * x/x = 4#

Multiply:

#(6(x + 3))/(x(x + 3)) + (7x)/(x(x + 3)) = 4#

Now we can add!

#(6(x + 3) + (7x))/(x(x + 3)) = 4#

Distribute and simplify:

#(6x + 18 + 7x)/(x^2 + 3x) = 4#

#(13x + 18)/(x^2 + 3x) = 4#

Let's multiply that denominator by itself, to both sides:

#(13x + 18)/(x^2 + 3x) * (x^2 + 3x) = 4 * (x^2 + 3x)#

#13x + 18 = 4x^2 + 12x#

Ahh, this is going to be a quadratic equation! Subtract both sides by #4x^2 + 12x#:

#13x + 18 - (4x^2 + 12x) = 4x^2 + 12x - (4x^2 + 12x)#

#-4x^2 + x + 18 = 0#

Let's complete the square, first dividing everything by #-4#:

#(-4x^2)/(-4) + x/(-4) + 18/(-4) = 0/(-4)#

#x^2 -1/4 (x) - 9/2 = 0#

Now, imagine a square #x^2# and a rectangle glued to its right with length #-1/4# and width #x#. Split that rectangle into half so that the other half glues to the bottom.

#x^2 + 2(-1/8)(x) - 9/2 = 0#

There's a spot on the bottom-right looking like a square can fit. The square would have a side length of #-1/8#, so its area is #(-1/8)^2#. Let's add it (to both sides, of course)!

#x^2 + 2(-1/8)(x) + (-1/8)^2 - 9/2 = 1/64#

Now look at the entire picture. All four shapes form a bigger square with side length #x - 1/8#, so the total area is #(x - 1/8)^2#. Replace all of the shapes with this (another way to look at this is using the rule #a^2 + 2ab + b^2 = (a + b)^2#).

#(x - 1/8)^2 - 9/2 = 1/64#

We're done completing the square; there's only one #x# now, so let's solve! Add both sides by #9/2#:

#(x - 1/8)^2 - 9/2 + 9/2 = 1/64 + 9/2#

#(x - 1/8)^2 = 1/64 + 288/64#

#(x - 1/8)^2 = 289/64#

Take the square root (but be careful! #3^2# and #(-3)^2# both equal #9#, so #sqrt(9) = ±3#):

#sqrt((x - 1/8)^2) = ±sqrt(289/64)#

Power of two and square root cancels out, and we can split the square root into the numerator and denominator:

#x - 1/8 = ±sqrt(289)/sqrt(64)#

#x - 1/8 = ±17/8#

Add #1/8# to both sides:

#x - 1/8 + 1/8 = 1/8 ± 17/8#

#x = 1/8 ± 17/8#

There are really two answers here: a positive and a negative one. We need both:

#x_1 = 1/8 + 17/8 = 18/8 = 9/4 = 2.25#

#x_2 = 1/8 - 17/8 = -16/8 = -2#

There we go! So the solutions are #x = {2.25, -2}#.

#6/x+7/(x+3)=4#

multiply both sides by #x#

#:.6(color(magenta)x/x)+(7(color(magenta)x))/(x+3)=4(color(magenta)x)#

#:.6+(7x)/(x+3)=4x#

multiply both sides by #color(magenta)(x+3#

#:.6color(magenta)((x+3))+7xcancel(color(magenta)((x+3)^1))/cancelcolor(magenta)color(magenta)(x+3)^1=4xcolor(magenta)((x+3)#

#:.6x+18+7x=4x^2+12x#

#:.4x^2+12x=6x+18+7x#

#:.4x^2+12x-6x-7x=18#

#:.4x^2-x-18=0#

#:.(4x-9)(x+2)=0#

#:.4x=9 or x=-2#

#:.color(magenta)(x=9/4 or x=-2#

~~~~~~~~~~~~~~~~

check:-

substitute# color(magenta)(x=-2#

#:.6/(color(magenta)((-2)))+7/((color(magenta)(-2)+3)=4#

#:.-3+7/1=4#

#:.-3+7=4#

#:.color(magenta)(4=4#

substitute# color(magenta)(x=9/4#

#:.6/(color(magenta)(9/4))+7/(color(magenta)(9/4)+3)=4#

#:.(cancel6^color(magenta)2/1xx4/cancel9^color(magenta)3)+7/((2 1/4)+3)=4#

#:.8/3+7/(5 1/4)=4#

#:.8/3+(cancel7^color(magenta)1/1xx4/cancel21^color(magenta)3)=4#

#8/3+4/3=4#

#:.12/3=4#

#:.color(magenta)(4=4#