Solve the inequality $\frac{x-5}{3} - \frac{2x-1}{2} \le -1$. Express as a single fraction in its simplest form $\frac{1}{x^2-4} - \frac{1}{x^2 - x - 6}$. Ally and Betty buy some pens and notebooks from the same shop.
Ally buys 3 pens and 2 notebooks for $4.80.
Betty buys 5 pens and 4 notebooks for $9. Form a pair of simultaneous equations to represent this information. Answer Solve the simultaneous equations to find the cost of a pen and the cost of a notebook. Answer Cost of pen = $ ... Cost of notebook = $ ... $p$ is inversely proportional to $q^2$. When $q$ is decreased by 50%, find the percentage change in $p$. Answer ... %

1. Clear the fractions by multiplying every term by the LCM of 3 and 2.

$\frac{x-5}{3}-\frac{2x-1}{2}\leq -1$
Multiply throughout by $6$:
$2(x-5)-3(2x-1)\leq -6$

2. Expand and simplify the linear inequality.

$2x-10-6x+3\leq -6$
$-4x-7\leq -6$

3. Solve for $x$, reversing the inequality sign when dividing by a negative number.

$-4x\leq 1$
Divide by $-4$ and reverse the inequality sign:
$x\geq -\frac14$

1. Factorise the two quadratic denominators.

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

$x^2-x-6=(x-3)(x+2)$

So
$\frac{1}{x^2-4}-\frac{1}{x^2-x-6}=\frac{1}{(x-2)(x+2)}-\frac{1}{(x-3)(x+2)}.$

2. Use the lowest common denominator to write as one fraction.

The lowest common denominator is $(x-2)(x+2)(x-3)$.

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

3. Subtract the numerators and cancel any common factors.

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

1. Define variables for the unknown costs and form the two equations from the given purchases.

Let $x$ be the cost of a pen and $y$ be the cost of a notebook.

Ally: $3$ pens and $2$ notebooks cost $4.80, so
$3x+2y=4.80$
Betty: $5$ pens and $4$ notebooks cost $9, so
$5x+4y=9$

1. Use elimination to find the cost of a pen.

From part (c)(i),
$3x+2y=4.80 \quad \text{and} \quad 5x+4y=9$
Multiply the first equation by 2:
$6x+4y=9.60$
Subtract the second equation:
$6x+4y-(5x+4y)=9.60-9$
$x=0.60$

2. Substitute the pen cost into one equation to find the cost of a notebook.

Substitute $x=0.60$ into $3x+2y=4.80$:
$3(0.60)+2y=4.80$
$1.80+2y=4.80$
$2y=3.00$
$y=1.50$

1. Write the inverse proportion relationship between $p$ and $q^2$.

Since $p$ is inversely proportional to $q^2$,

$p=\frac{k}{q^2}$

where $k$ is a constant.

2. Express the new value of $q$ after a 50% decrease and find the new value of $p$.

When $q$ is decreased by $50\%$,

$q_{\text{new}}=50\%\text{ of }q=\frac{1}{2}q$

So

$p_{\text{new}}=\frac{k}{\left(\frac{1}{2}q\right)^2}$

$=\frac{k}{\frac{1}{4}q^2}=\frac{4k}{q^2}=4p$

3. Calculate the percentage change in $p$.

Since $p_{\text{new}}=4p$,

increase in $p$ is

$4p-p=3p$

Percentage increase

$=\frac{3p}{p}\times 100\%=300\%$