Using a separate diagram for each part, represent on the number line the solution set of

1. Expand the left-hand side and simplify the inequality.

$3(5-x)>x+13$

$15-3x>x+13$

2. Collect like terms and solve for $x$.

$15-3x>x+13$

$15-13>x+3x$

$2>4x$

$x<\dfrac{1}{2}$

3. Represent the solution set on the number line.

Since $x<\dfrac{1}{2}$, draw an open circle at $\dfrac{1}{2}$ and shade the number line to the left.

1. Clear fractions and collect all terms on one side.

$x^{2}-\dfrac{1}{2}x<\dfrac{1}{4}(3-2x)$

Multiply throughout by $4$:

$4x^{2}-2x<3-2x$

Add $2x$ to both sides:

$4x^{2}<3$

$4x^{2}-3<0$

2. Factorise the quadratic expression using difference of squares.

$4x^{2}-3<0$

$(2x)^{2}-(\sqrt{3})^{2}<0$

$(2x-\sqrt{3})(2x+\sqrt{3})<0$

3. Determine the interval where the quadratic expression is negative and represent it on the number line.

Critical values:

$2x-\sqrt{3}=0 \Rightarrow x=\dfrac{\sqrt{3}}{2}$

$2x+\sqrt{3}=0 \Rightarrow x=-\dfrac{\sqrt{3}}{2}$

Since $4x^{2}-3<0$, the quadratic is below $0$ between the two roots.

Therefore,

$-\dfrac{\sqrt{3}}{2}<x<\dfrac{\sqrt{3}}{2}$

1. Find the overlap of the two solution sets from parts (i) and (ii).

From (i), $x<\dfrac{1}{2}$.

From (ii), $-\dfrac{1}{2}<x<\dfrac{3}{2}$.

The values satisfying both are the common values:

$-\dfrac{1}{2}<x<\dfrac{1}{2}$