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)
$\displaystyle \frac{x}{6} - \frac{x}{8} = \frac{1}{12}$
Multiply every term by 24
$\displaystyle 4x - 3x = 2$
$\displaystyle x = 2$
$\displaystyle \frac{4}{a-7} = \frac{-2a}{a+3}$
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 $\displaystyle x = 2$
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 $\displaystyle x = -1$