Solve the inequality $\frac{-3x+4}{2} < \frac{5-x}{7}$. Answer ... Simplify $\left(\dfrac{64a^{6}}{b^{-15}}\right)^{-\tfrac{1}{3}}$, expressing your answer in positive index notation. Answer ... Express $\dfrac{2}{2x-3} + \dfrac{7}{2x^{2} + x - 6}$ as a single fraction in its simplest form. Answer ... Simplify $\dfrac{4u^{2} - v^{2}}{6uv - 3v^{2}}$. Answer ...

1. Clear the fractions by multiplying both sides by the positive common denominator.

$\frac{-3x+4}{2} < \frac{5-x}{7}$
Multiply both sides by $14$:
$7(-3x+4) < 2(5-x)$

2. Expand and solve the linear inequality.

$-21x+28 < 10-2x$
$28 < 10+19x$
$18 < 19x$
$x > \frac{18}{19}$

1. Rewrite division by a negative power and simplify the expression inside the brackets.

$\left(\dfrac{64a^{6}}{b^{-15}}\right)^{-\frac13}=\left(64a^6b^{15}\right)^{-\frac13}$

2. Apply the power $-\frac13$ to each factor and express the answer using positive indices.

$\left(64a^6b^{15}\right)^{-\frac13}=64^{-\frac13}a^{-2}b^{-5}$

Since $64^{\frac13}=4$,

$64^{-\frac13}=\dfrac14$.

So

$64^{-\frac13}a^{-2}b^{-5}=\dfrac{1}{4a^2b^5}$

1. Factorise the quadratic denominator.

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

So
$\frac{2}{2x-3}+\frac{7}{2x^{2}+x-6}=\frac{2}{2x-3}+\frac{7}{(2x-3)(x+2)}.$

2. Write both fractions with the common denominator.

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

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

Hence
$\frac{2}{2x-3}+\frac{7}{(2x-3)(x+2)}=\frac{2(x+2)+7}{(2x-3)(x+2)}.$

3. Expand and simplify the numerator.

$\frac{2(x+2)+7}{(2x-3)(x+2)}=\frac{2x+4+7}{(2x-3)(x+2)}=\frac{2x+11}{(2x-3)(x+2)}.$

1. Factorise the numerator and denominator.

$\frac{4u^{2}-v^{2}}{6uv-3v^{2}}=\frac{(2u+v)(2u-v)}{3v(2u-v)}$

2. Cancel the common algebraic factor.

$\frac{(2u+v)(2u-v)}{3v(2u-v)}=\frac{2u+v}{3v}$