Inequalities

(a) The open circle at $-1$ means $-1$ is not included, and the arrow points right: $x > -1$.

(b) Filled circle at $-3$ (included) and open circle at 4 (not included), with a line joining them.

(a) $n$ is less than 5, so the greatest integer is $4$.

(b) Filled circle at $-4$ (it is included), empty circle at $1$ (not included), and a line joining them.

(c) $\frac{2}{5}g - 4 < 6$

$\frac{2}{5}g < 10$

$2g < 50$

$g < 25$

(a) $\le 4$ includes 4, so the circle at 4 should be solid. The line should start with an open circle at $-3$, but Jenna started at $-2$.

(b) $5y - 7 < 16$ gives $5y < 23$, so $y < 4.6$.

The greatest integer is 4.

$7x - 27 < 8$

Add 27: $7x < 35$

Divide by 7: $x < 5$

The filled circle at $-3$ means $-3$ is included: $p \ge -3$

The open circle at $1$ means $1$ is not included: $p < 1$

Together: $-3 \le p < 1$

(a) $14n - 11n > 6$, so $3n > 6$ and $n > 2$.

(b) Subtract 3 from every part: $-2 - 3 < x \le 4 - 3$, so $-5 < x \le 1$.

Draw an open circle at $-5$ (not included), a filled circle at $1$ (included) and join them.

(a) $x < 4$ does not include 4, so draw an open circle at 4 and an arrow pointing left.

(b) $y$ is more than 3 and at most 7: $4, 5, 6, 7$

(c) $3x + 5 \geqslant x + 17$

$2x + 5 \geqslant 17$

$2x \geqslant 12$, so $x \geqslant 6$

(a) $\frac{5x}{2} + 3 > 18$

$\frac{5x}{2} > 15$, so $5x > 30$ and $x > 6$

(b) Need two numbers that multiply to 9 and add to 10: 9 and 1.

$(x + 9)(x + 1)$

Possible x: 4, 5, 6, 7. Possible y: 5, 6, 7, 8, 9

$y = 14 - x$: x = 4 gives y = 10 (not allowed); x = 5, 6, 7 give y = 9, 8, 7 (all allowed)

So x = 5, 6 or 7

$x + 11 \le 5 - \frac{1}{2}x$

$\frac{3}{2}x \le -6$

$x \le -4$

Draw the four lines: $x = 2$ (vertical), $y = -3$ (horizontal), $y = 2x + 1$ (through $(0,1)$, gradient 2) and $3x + 2y = 6$ (through $(0,3)$ and $(2,0)$).

The region is left of $x=2$, above $y=-3$, below $y=2x+1$ and below $3x+2y=6$.

The region is a quadrilateral with corners $(-2,-3)$, $(\tfrac47, \tfrac{15}{7})$, $(2, 0)$ and $(2,-3)$. Shade it and label it R.

Critical values: $4x - 3 = 0 \Rightarrow x = \frac34$ and $x + 5 = 0 \Rightarrow x = -5$.

The product is negative between the roots (the graph is a U shape below the axis there).

So $-5 < x < \frac34$.

(a) $6 \times (-4) = -24$; use $-8$ and $3$: $6x^2 - 8x + 3x - 4 = 2x(3x - 4) + (3x - 4) = (2x + 1)(3x - 4)$

(b) Critical values: $x = -\frac{1}{2}$ and $x = \frac{4}{3}$

The graph is a U shape, so it is below zero between the roots: $-\frac{1}{2} < x < \frac{4}{3}$

Horizontal line: $y = 6$, region below, so $y \le 6$.

Vertical line: $x = -3$, region to the right, so $x \ge -3$.

Steep line through $(0, 6)$ and $(-2, 0)$: gradient 3, so $y = 3x + 6$; region above, $y \ge 3x + 6$.

Shallow line through $(0, 1)$ and $(2, 0)$: gradient $-\frac{1}{2}$; region above, $y \ge -\frac{1}{2}x + 1$.

Draw $x = 2y + 4$, i.e. $y = \frac{1}{2}x - 2$ (through (0, −2) and (4, 0)).

Draw $x = 3$ (vertical) and $y = 6 - 3x$ (through (0, 6) and (2, 0)).

$2y + 4 < x$ means below the first line; $x < 3$ means left of $x = 3$; $y < 6 - 3x$ means below the third line.

The two sloping lines cross at about $(2.3, -0.9)$ and $y = 6 - 3x$ meets $x = 3$ at $(3, -3)$. Shade the part of the grid below both sloping lines and left of $x = 3$, and label it R.

(a) Draw $x = 2$ (vertical), $y = x + 3$ (through $(0,3)$ and $(2,5)$) and $2x + 3y = 6$ (through $(0,2)$ and $(3,0)$).

The region is between $x = 0$ and $x = 2$, below $y = x + 3$ and above $2x + 3y = 6$. Label it R.

(b) At $(2, 4)$: $y = 4 \le 4x = 8$, $y = 4 \ge \frac{1}{2}x = 1$, and $x + y = 6 \le 6$. All are satisfied because the point is on the boundary, so Geoffrey is wrong.

Draw $x + y = 4$ (through $(0, 4)$ and $(4, 0)$), $y = x - 1$ (through $(0, -1)$ and $(1, 0)$) and $y = 3x$ (through $(0, 0)$ and $(1, 3)$). All are strict, so dashed lines are fine.

Test $(1, 1)$: $2 < 4$, $1 > 0$, $1 < 3$. All true, so the region containing $(1, 1)$ is wanted.

R is the triangle with corners $(1, 3)$, $(2.5, 1.5)$ and $(-0.5, -1.5)$.

$x^2 - 49 > 0$: $(x + 7)(x - 7) > 0$, so $x < -7$ or $x > 7$

$5x^2 - 31x - 72 > 0$: $(5x + 9)(x - 8) > 0$, so $x < -1.8$ or $x > 8$

Both true: $x < -7$ or $x > 8$

(a) Area $= ab\sin C = (2x - 1)(10 - x) \times \sin 150° = \frac{1}{2}(2x - 1)(10 - x)$

$= \frac{1}{2}(-2x^2 + 21x - 10)$

So $\frac{1}{2}(-2x^2 + 21x - 10) > 15 \Rightarrow -2x^2 + 21x - 10 > 30 \Rightarrow 2x^2 - 21x + 40 < 0$

(b) $(2x - 5)(x - 8) < 0$, critical values $x = 2.5$ and $x = 8$.

The U-shaped graph is below zero between the roots, so $2.5 < x < 8$.

Multiply by 4: $88 < m^2 + 7 < 128$

Subtract 7: $81 < m^2 < 121$

So $9 < |m| < 11$.

$9 < m < 11$ or $-11 < m < -9$

$4x^2 - 25 = (2x + 5)(2x - 5) < 0$ gives $-2.5 < x < 2.5$

$12 - 5x - 3x^2 > 0$ is the same as $3x^2 + 5x - 12 < 0$, i.e. $(3x - 4)(x + 3) < 0$, giving $-3 < x < \frac{4}{3}$

Both must hold, so $-2.5 < x < \frac{4}{3}$

$3n + 2 \le 14 \Rightarrow 3n \le 12 \Rightarrow n \le 4$

$n^2 + 5$ is always positive, so $\frac{6n}{n^2 + 5} > 1 \Rightarrow 6n > n^2 + 5 \Rightarrow n^2 - 6n + 5 < 0$

$(n - 1)(n - 5) < 0 \Rightarrow 1 < n < 5$

Both together: $1 < n \le 4$, so $n = 2, 3, 4$