You should’ve put a picture to make it easier for us
The first step for solving this equation is to determine the defined range.

, x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:

= 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/-
![\sqrt[4]{5}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B5%7D)
Separate the solutions.
x =
![\sqrt[4]{5}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B5%7D)
, x ≠ 1
x = -
Check if the solution is in the defined range.
x =
x = -
This means that the final solution to your question are the following:
x =
x = -
Let me know if you have any further questions.
:)
Answer:
x = 4, y = 2
Step-by-step explanation:
Start by multiplying the first equation by 2:
2x + 2y = 12 --> 4x + 4y = 24
Subtract the second from the first:
4x + 4y = 24
- 5x + 4y = 28
4x - 5x = -x
4y = 4y = 0
24 - 28 = -4
so you end with -x + 0 = -4
Solve for x to get x = 4
Plug x = 4 back into 2x + 2y = 12 to find y.
2(4) + 2y = 12
8 + 2y = 12
2y = 4
y = 2

Setting

, you have

. Then the integral becomes




Now,

in general. But since we want our substitution

to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means

, which implies that

, or equivalently that

. Over this domain,

, so

.
Long story short, this allows us to go from

to


Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

Then integrate term-by-term to get


Now undo the substitution to get the antiderivative back in terms of

.

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to