how would you guys solve it?

[loud glade]

how would you guys solve it and similar rational equations?

[left quarry]

By decomposing the partial fractions.

Multiply both sides by 3x+2 and 2x+1

[loud glade]

but how do you figure one what to do in any rational equation like this? i have seen many are solved differently and not one method can satisfy all of them

[left quarry]

It depends on the rational equation. This one can decompose nicely

You should actually multiply by

(2x+1)(5x+4)

Then 3x+2

Any order doesn’t matter

Just remove the denominators

Get a fraction, say ax+b/(f(x))

Then solve the top

Check that f(solution) != 0

Then you have finished

[loud glade]

that makes thing even more complicated

@left quarry can you do it on a paper and show me?

[icy trench]

Why don't you think one method can solve all rational equations?

[left quarry]

No, it removes the first two radicals

[left quarry]

Since he's weird ig

One method can solve all rational equations

Literally called decomposing the fractions

Let mw do it

$\frac{4}{2x+1} + \frac{9}{3x+2}= \frac{25}{5x+4}$

[carmine tangle]

Lucifer

[left quarry]

Next we will multiply by $(2x+1)(3x+2)$

[carmine tangle]

Lucifer

[left quarry]

$(2x+1)(3x+2)\left(\frac{4}{2x+1} + \frac{9}{3x+2}\right)=(2x+1)(3x+2)\frac{25}{5x+4}$

[carmine tangle]

Lucifer

[left quarry]

Reduce LHS

$4(3x+2) + 9(2x+1)=(2x+1)(3x+2)\frac{25}{5x+4}$

[carmine tangle]

Lucifer

[left quarry]

@loud glade this should be trivial to solve now

Expanding the right hand side:

$(2x+1)(3x+2)=6x^2+7x+2$

[carmine tangle]

Lucifer

[left quarry]

$\frac{25(6x^2+7x+2)}{5x+4}$

[carmine tangle]

Lucifer

[left quarry]

We can then simplify:

There are two routes now

Multiply both sides by 5x+4

$(5x+4)(4(3x+2) + 9(2x+1))=25(6x^2+7x+2)$

[carmine tangle]

Lucifer

[left quarry]

Then just expand this to get two quadratics:

$150x^2 + 205x + 68$

[carmine tangle]

Lucifer

[left quarry]

Then just solve this:

$150x^2 + 205x + 68=25(6x^2+7x+2)$

[carmine tangle]

Lucifer