Solve the inequality $\frac{4x+1}{3} > \frac{3-2x}{5}$. Simplify $\left(\dfrac{16a^{12}}{b^8}\right)^{-\tfrac{1}{4}}$, leaving your answer in positive indices. Express $\dfrac{x}{(5-2x)^2} - \dfrac{3}{2x-5}$ as a single fraction in its simplest form. Solve these simultaneous equations. $7x + 6y = 33$ $5x - 4y = 7$ Answer (d) x = ... y = ... Simplify $\dfrac{25x^2-16}{15x^2+7x-4}$. Answer (e) ...

1. Clear the denominators without changing the inequality sign.

$\dfrac{4x+1}{3}>\dfrac{3-2x}{5}$

Multiply both sides by $15$:

$5(4x+1)>3(3-2x)$

2. Expand, collect like terms and isolate $x$.

$20x+5>9-6x$

$26x>4$

$x>\dfrac{4}{26}$

$x>\dfrac{2}{13}$

1. Change the negative index to a reciprocal with a positive fractional index.

$\left(\dfrac{16a^{12}}{b^8}\right)^{-\frac14}=\left(\dfrac{b^8}{16a^{12}}\right)^{\frac14}$

2. Apply the power $\frac14$ to each factor and simplify using positive indices.

$\left(\dfrac{b^8}{16a^{12}}\right)^{\frac14}=\dfrac{(b^8)^{\frac14}}{16^{\frac14}(a^{12})^{\frac14}}$
$=\dfrac{b^2}{2a^3}$

1. Rewrite the squared denominator in terms of $2x-5$.

Since $5-2x=-(2x-5)$,
$(5-2x)^2=(2x-5)^2.$
So
$\frac{x}{(5-2x)^2}-\frac{3}{2x-5}=\frac{x}{(2x-5)^2}-\frac{3}{2x-5}.$

2. Use the common denominator and combine into a single fraction.

The common denominator is $(2x-5)^2$.
\frac{x}{(2x-5)^2}-\frac{3}{2x-5}=rac{x}{(2x-5)^2}-\frac{3(2x-5)}{(2x-5)^2}
$=\frac{x-3(2x-5)}{(2x-5)^2}$
$=\frac{x-6x+15}{(2x-5)^2}$
$=\frac{15-5x}{(2x-5)^2}.$

1. Make the coefficients of $y$ equal in magnitude.

$7x+6y=33 \quad \cdots (1)$
$5x-4y=7 \quad \cdots (2)$
Multiply $(1)$ by $2$:
$14x+12y=66 \quad \cdots (3)$
Multiply $(2)$ by $3$:
$15x-12y=21 \quad \cdots (4)$

2. Eliminate $y$ and solve for $x$.

Add $(3)$ and $(4)$:
$14x+12y+15x-12y=66+21$
$29x=87$
$x=3$

3. Substitute $x=3$ into one original equation to find $y$.

Substitute $x=3$ into $(2)$:
$5(3)-4y=7$
$15-4y=7$
$-4y=-8$
$y=2$

1. Factorise the numerator using difference of squares.

$25x^2-16=(5x)^2-4^2=(5x-4)(5x+4)$

2. Factorise the quadratic denominator.

$15x^2+7x-4=(5x+4)(3x-1)$

3. Cancel the common factor and write the simplified fraction.

$\frac{25x^2-16}{15x^2+7x-4}=\frac{(5x-4)(5x+4)}{(5x+4)(3x-1)}=\frac{5x-4}{3x-1}$