Adam has some 50-cent and 20-cent coins in his wallet. Given that he has a total of 73 coins adding up to a value of $28.10, form 2 simultaneous equations to find the number of 50-cent and 20-cent coins he has. Answer He has ... 50-cent coins and ... 20-cent coins Express $\dfrac{4x}{4x^2-25} + \dfrac{1}{5-2x}$ as a single fraction in its simplest form.

1. Define the unknowns clearly.

Let $x$ be the number of 50-cent coins and $y$ be the number of 20-cent coins.

2. Form the equation for the total number of coins.

Since there are $73$ coins altogether,

$x+y=73.$

3. Form the equation for the total value of the coins.

Using dollars, a 50-cent coin is worth $0.50$ dollars and a 20-cent coin is worth $0.20$ dollars.

$0.50x+0.20y=28.10.$

Equivalently, multiplying by $10$,

$5x+2y=281.$

4. Solve the simultaneous equations.

From $x+y=73$,

$y=73-x.$

Substitute into $5x+2y=281$:

$5x+2(73-x)=281$
$5x+146-2x=281$
$3x=135$
$x=45.$

Then

$y=73-45=28.$

1. Factorise the quadratic denominator using difference of squares.

$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5)$
So
$\frac{4x}{4x^2-25}+\frac{1}{5-2x}=\frac{4x}{(2x-5)(2x+5)}+\frac{1}{5-2x}.$

2. Rewrite the second denominator so that it matches a factor of the common denominator.

Since
$5-2x=-(2x-5),$
$\frac{1}{5-2x}=-\frac{1}{2x-5}.$

3. Combine the fractions over the common denominator and simplify the numerator.

$\frac{4x}{(2x-5)(2x+5)}-\frac{1}{2x-5}$
$=\frac{4x}{(2x-5)(2x+5)}-\frac{2x+5}{(2x-5)(2x+5)}$
$=\frac{4x-(2x+5)}{(2x-5)(2x+5)}$
$=\frac{2x-5}{(2x-5)(2x+5)}$
$=\frac{1}{2x+5}.$