Show that $\dfrac{2x^{2}-x-19}{x^{2}+3x+2}-1=\dfrac{x^{2}-4x-21}{x^{2}+3x+2}$. Hence, without using a calculator, solve the inequality $\dfrac{2x^{2}-x-19}{x^{2}+3x+2}>1.$

1. Express both terms with a common denominator.

$\frac{2x^{2}-x-19}{x^{2}+3x+2}-1 = \frac{2x^{2}-x-19}{x^{2}+3x+2}-\frac{x^{2}+3x+2}{x^{2}+3x+2}$

2. Combine the numerators and simplify to obtain the required expression.

1. Rewrite the inequality in the form $\frac{\mathrm{f}(x)}{\mathrm{g}(x)} > 0$ using the identity from part [0].

$\dfrac{2x^{2}-x-19}{x^{2}+3x+2} > 1$
$\dfrac{2x^{2}-x-19}{x^{2}+3x+2} - 1 > 0$
Using the result from part [0]:
$\dfrac{x^{2}-4x-21}{x^{2}+3x+2} > 0$

2. Factorise the quadratic expressions in both the numerator and the denominator.

$\dfrac{(x-7)(x+3)}{(x+1)(x+2)} > 0$

3. Identify the critical values of the rational expression.

The critical values occur where the numerator or denominator is zero:
$x - 7 = 0 \implies x = 7$
$x + 3 = 0 \implies x = -3$
$x + 1 = 0 \implies x = -1$
$x + 2 = 0 \implies x = -2$

4. Determine the sign of the fraction in each interval to find the solution.

Testing the intervals defined by the critical values $x = -3$, $x = -2$, $x = -1$, and $x = 7$:

• For $x < -3$: the fraction is positive ($> 0$)
• For $-3 < x < -2$: the fraction is negative ($< 0$)
• For $-2 < x < -1$: the fraction is positive ($> 0$)
• For $-1 < x < 7$: the fraction is negative ($< 0$)
• For $x > 7$: the fraction is positive ($> 0$)

The inequality requires the expression to be strictly greater than $0$, so we take the positive intervals.

$x < -3 \text{ or } -2 < x < -1 \text{ or } x > 7$