解方程 $x/5 + (2x-3)/4 = -6$。 答：x = ... 化简 $2(p-3q)/(10r) + (3q-p)^2/r$。 答：... 解不等式 $x+1 \le \dfrac{7x-2}{4} < \dfrac{5x+19}{5}$。 答：... 化简 $\dfrac{2y^2+7y-9}{4y^2-81}$。 答：...

1. Clear the fractions using the LCM of the denominators 5 and 4.

$\frac{x}{5}+\frac{2x-3}{4}=-6$
Multiply throughout by $20$:
$4x+5(2x-3)=-120$

2. Expand, collect like terms and solve for $x$.

$4x+10x-15=-120$
$14x-15=-120$
$14x=-105$
$x=\frac{-105}{14}=-\frac{15}{2}$

1. Rewrite the expression so both terms use the common factor $p-3q$.

$\dfrac{2(p-3q)}{10r}+\dfrac{(3q-p)^2}{r}=\dfrac{p-3q}{5r}+\dfrac{(p-3q)^2}{r}$

2. Combine over a common denominator and factorise the numerator.

$\dfrac{p-3q}{5r}+\dfrac{(p-3q)^2}{r}=\dfrac{p-3q+5(p-3q)^2}{5r}$

$=\dfrac{(p-3q)\{1+5(p-3q)\}}{5r}$

$=\dfrac{(p-3q)(5p-15q+1)}{5r}$

1. Split the compound inequality into two separate inequalities.

$x+1 \leq \frac{7x-2}{4} \quad \text{and} \quad \frac{7x-2}{4} < \frac{5x+19}{5}$

2. Solve the left-hand inequality.

$x+1 \leq \frac{7x-2}{4}$
Multiply by $4$:
$4x+4 \leq 7x-2$
$6 \leq 3x$
$x \geq 2$

3. Solve the right-hand inequality.

$\frac{7x-2}{4} < \frac{5x+19}{5}$
Multiply by $20$:
$5(7x-2) < 4(5x+19)$
$35x-10 < 20x+76$
$15x < 86$
$x < \frac{86}{15}$

4. Combine the two solution ranges.

From the two inequalities,
$x \geq 2 \quad \text{and} \quad x < \frac{86}{15}.$
Therefore,
$2 \leq x < \frac{86}{15}.$

1. Factorise the numerator.

$2y^2+7y-9=(2y-1)(y+9)$

2. Factorise the denominator using difference of squares.

$4y^2-81=(2y)^2-9^2=(2y-9)(2y+9)$

3. Write the fraction in factorised form and cancel the common factor.

$\frac{2y^2+7y-9}{4y^2-81}=\frac{(2y-1)(y+9)}{(2y-9)(2y+9)}$

There is no common factor to cancel, so the expression is already in simplest factorised form.