Simplify $\dfrac{25m^2-4}{4-4m-15m^2}$. Given that $r = 3 - \sqrt{\dfrac{p^2}{4} - q}$, evaluate $r$ when $p = -8$ and $q = 7$. express $p$ in terms of $q$ and $r$. The value of a 2-digit number can be written as 10x + y where x is the tens digit of the number and y is its units digit. When the digits x and y are reversed, the value is increased by 27. Show that $y - x = 3$. The sum of the original number and the number with reverse digits is 121. Show that $x + y = 11$. Hence, find the original number. Answer ...

1. Factorise the numerator using difference of two squares.

$25m^2-4=(5m)^2-2^2=(5m-2)(5m+2)$

2. Factorise the quadratic denominator.

$4-4m-15m^2=-(15m^2+4m-4)$
$=-(5m-2)(3m+2)$
Equivalently,
$4-4m-15m^2=(2-5m)(3m+2).$

3. Cancel the common factor to simplify the algebraic fraction.

$\frac{25m^2-4}{4-4m-15m^2}=\frac{(5m-2)(5m+2)}{(2-5m)(3m+2)}$
Since $2-5m=-(5m-2)$,
$\frac{(5m-2)(5m+2)}{-(5m-2)(3m+2)}=-\frac{5m+2}{3m+2}=\frac{-5m-2}{3m+2}.$
This may also be written as
$\frac{2-5m}{3m+2}.$

1. Substitute $p=-8$ and $q=7$ into the formula and evaluate.

$r=3-\sqrt{\frac{(-8)^2}{4}-7}$
$=3-\sqrt{\frac{64}{4}-7}$
$=3-\sqrt{16-7}$
$=3-\sqrt{9}=3-3=0$

1. Isolate the square root term.

$r=3-\sqrt{\dfrac{p^2}{4}-q}$

$\sqrt{\dfrac{p^2}{4}-q}=3-r$

2. Square both sides to remove the square root.

$\dfrac{p^2}{4}-q=(3-r)^2$

3. Make $p$ the subject of the formula.

$\dfrac{p^2}{4}=q+(3-r)^2$

$p^2=4\left[q+(3-r)^2\right]$

$p=\pm\sqrt{4\left[q+(3-r)^2\right]}$

$p=\pm 2\sqrt{q+(3-r)^2}$

1. Write the value of the reversed 2-digit number and form the equation.

Original number $=10x+y$.

Reversed number $=10y+x$.

Since the value is increased by $27$ when reversed,
$10y+x=10x+y+27.$

2. Simplify the equation to obtain the required result.

$10y+x=10x+y+27$
$9y-9x=27$
$y-x=3.$

1. Form the equation for the sum of the original number and the reversed number.

Original number $=10x+y$

Reversed number $=10y+x$

Given their sum is $121$,
$10x+y+10y+x=121.$

2. Simplify the equation to obtain the required result.

$11x+11y=121$
$11(x+y)=121$
$x+y=11$

1. Use the two earlier results as simultaneous equations.

From parts (c)(i) and (c)(ii),
$y-x=3 \quad \text{and} \quad x+y=11.$

2. Solve for the two digits $x$ and $y$.

Add the equations:
$y-x+x+y=3+11$
$2y=14$
$y=7.$
Substitute into $x+y=11$:
$x+7=11$
$x=4.$

3. Find the value of the original 2-digit number.

The original number is $10x+y$.
$10x+y=10(4)+7=47.$