1. ## solve the equations

im soooo lost any help would be amazing
1.
(x) (x) (1)
_ - _ = ___
(6) (8) (12)

2.

(4) (-2a)
___ = ___

(a-7) (a+3)

3.

(x) (5) (1)
___ + ___ = ____
(x+1) (x) (x^2+x)

2. i dont know why it does not look right but the () are grouped so its like the first groupe is part of the first fraction and so on

im soooo lost any help would be amazing
1.
(x) (x) (1)
_ - _ = ___
(6) (8) (12)
$\displaystyle \frac{x}{6} - \frac{x}{8} = \frac{1}{12}$

Multiply every term by 24

$\displaystyle 4x - 3x = 2$

$\displaystyle x = 2$

2.

(4) (-2a)
___ = ___

(a-7) (a+3)
$\displaystyle \frac{4}{a-7} = \frac{-2a}{a+3}$

Multiply every term with $\displaystyle (a - 7)(a + 3)$

$\displaystyle 4(a+3) = (-2a)(a-7)$

$\displaystyle 4a + 12 = -2a^2 + 14a$

$\displaystyle 2a^2 - 10 + 12 = 0$

$\displaystyle 2(a^2 - 5 + 6) = 0$

$\displaystyle 2(a - 3)(a - 2) = 0$

Thus $\displaystyle x = 3$ or $\displaystyle x = 2$

3.

(x) (5) (1)
___ + ___ = ____
(x+1) (x) (x^2+x)
First notice that $\displaystyle (x^2 + x) = (x)(x + 1)$. Always factorise!

Now multiply every term by $\displaystyle (x)(x + 1)$.

$\displaystyle (x)(x) + (5)(x+1) = 1$

$\displaystyle x^2 + 5x + 5 - 1 = 0$

$\displaystyle x^2 + 5x + 4 = 0$

$\displaystyle (x + 4)(x + 1) = 0$

Thus $\displaystyle x = -4$ or $\displaystyle x = -1$