一家超市在一月份以每包 $$x$ 的价格售出了其所有的卷筒卫生纸。一月份售出卷筒卫生纸的收入为 $5940。

1. Form an expression for the number of packs sold in January from revenue and price per pack.

Revenue in January = $5940, price per pack = $$x$.
$\text{Number of packs sold}=\frac{5940}{x}$

1. Form the February revenue expression from number of packs and price per pack.

Number of packs in February $=\dfrac{5940}{x}+600$.

Price per pack in February $=x+0.50$ dollars.

Total amount received in February
$=\left(\dfrac{5940}{x}+600\right)(x+0.50).$

1. Write an equation showing that February revenue is \$3870 more than January revenue.

From part (b), February revenue is
$\left(\dfrac{5940}{x}+600\right)(x+0.50).$
January revenue is $5940.

Since February revenue is $3870 more than January revenue,
$\left(\dfrac{5940}{x}+600\right)(x+0.50)-5940=3870.$

2. Expand and simplify the equation.

$\left(\dfrac{5940}{x}+600\right)(x+0.50)-5940=3870$
$5940+\dfrac{2970}{x}+600x+300-5940=3870$
$600x+\dfrac{2970}{x}+300=3870$

3. Clear the denominator and reduce to the required quadratic equation.

$600x+\dfrac{2970}{x}+300=3870$
$600x+\dfrac{2970}{x}=3570$
Multiplying by $x$,
$600x^2+2970=3570x$
$600x^2-3570x+2970=0$
Divide throughout by $30$:
$20x^2-119x+99=0.$

1. Factorise the quadratic expression.

$20x^2-119x+99=0$

Find two numbers with product $20\times 99=1980$ and sum $-119$: $-44$ and $-75$.

$20x^2-44x-75x+99=0$

$4x(5x-11)-9(5x-11)=0$

$(5x-11)(4x-9)=0$

2. Use the zero product property to find the two values of $x$.

$(5x-11)(4x-9)=0$

So,

$5x-11=0 \quad \text{or} \quad 4x-9=0$

$x=\frac{11}{5} \quad \text{or} \quad x=\frac{9}{4}$

1. Select the valid value of $x$ using the condition that each pack costs more than \$1.

From part (d), $x=\dfrac{9}{10}$ or $x=\dfrac{11}{2}$.

Since each pack was sold for more than $1, $x>1$.

Hence, $x=\dfrac{11}{2}=5.50$.

2. Find the number of packs sold in February.

Number of packs sold in February
$=\frac{5940}{x}+600$
$=\frac{5940}{5.5}+600$
$=1080+600$
$=1680$