Express as a single fraction in its simplest form $\frac{2}{7-3x} - \frac{1}{6-x}.$ It is given that $v = \frac{3-5w}{w+2} + 9.$ Find v when w = -6. Rearrange the formula to make w the subject. Solve the equation $\dfrac{5}{x-2} - \dfrac{3}{x^2-4} = \dfrac{1}{7}$. Give your solutions correct to two decimal places.

1. Use a common denominator to combine the two algebraic fractions.

$\frac{2}{7-3x}-\frac{1}{6-x}=\frac{2(6-x)-(7-3x)}{(7-3x)(6-x)}$

2. Expand and simplify the numerator.

$\frac{2(6-x)-(7-3x)}{(7-3x)(6-x)}=\frac{12-2x-7+3x}{(7-3x)(6-x)}=\frac{x+5}{(7-3x)(6-x)}$

1. Substitute $w=-6$ into the given formula and evaluate.

$v=\frac{3-5(-6)}{-6+2}+9$
$=\frac{3+30}{-4}+9$
$=\frac{33}{-4}+9$
$=-\frac{33}{4}+\frac{36}{4}=\frac{3}{4}$

1. Start from the formula and isolate the algebraic fraction.

$v=\frac{3-5w}{w+2}+9$
$v-9=\frac{3-5w}{w+2}$

2. Clear the denominator by multiplying both sides by $w+2$.

$(v-9)(w+2)=3-5w$
$vw+2v-9w-18=3-5w$

3. Collect the terms involving $w$, factorise, and divide to make $w$ the subject.

$vw-9w+5w=3+18-2v$
$vw-4w=21-2v$
$w(v-4)=21-2v$
$w=\frac{21-2v}{v-4}$

1. State the restriction and factorise the quadratic denominator.

Since the denominators must not be zero,
$x\neq 2$ and $x\neq -2$.

$x^2-4=(x-2)(x+2)$

So
$\frac{5}{x-2}-\frac{3}{(x-2)(x+2)}=\frac{1}{7}.$

2. Clear the denominators using the common denominator.

Multiply throughout by $7(x-2)(x+2)$:

$35(x+2)-21=(x-2)(x+2).$

3. Expand and rearrange into standard quadratic form.

$35x+70-21=x^2-4$

$35x+49=x^2-4$

$x^2-35x-53=0$

4. Solve the quadratic equation.

Using the quadratic formula on $x^2-35x-53=0$,

$x=\frac{-(-35)\pm\sqrt{(-35)^2-4(1)(-53)}}{2(1)}$

$x=\frac{35\pm\sqrt{1437}}{2}.$

5. Evaluate, round to two decimal places, and check the solutions are valid.

$x=\frac{35+\sqrt{1437}}{2}=36.437\ldots$

$x=\frac{35-\sqrt{1437}}{2}=-1.437\ldots$

Both are not equal to $2$ or $-2$, so both are valid.

Correct to two decimal places:

$x=36.44\quad \text{or}\quad x=-1.44.$