Find the critical values: x^2 - 9 = 0 \;\; (x - 3) (x + 3) = 0 \;\; x = -3 or x = 3 Sketch the parabola and pick the side you need: y = x^2 - 9 State the solution: -3 < x < 3 Answer: -3 < x < 3

The region R satisfies y < x + 2, y ≥ -1 and x ≤ 3. Which statement about the three boundary lines of R is correct?

Write down the inequalities that define R: y < x + 2, y ≥ -1, x ≤ 3 Deal with y < x + 2: boundary y = x + 2 (dashed) Deal with y ≥ -1: boundary y = -1 (solid) Deal with x ≤ 3: boundary x = 3 (solid) Sketch the region R: y < x + 2, y ≥ -1, x ≤ 3 State which boundaries are solid: y < x + 2 arrow dashed, y ≥ -1 arrow solid, x ≤ 3 arrow solid Answer: y < x + 2 arrow dashed, y ≥ -1 arrow solid, x ≤ 3 arrow solid

GCSE Quadratic & 2variable inequalities Practice Questions

Free GCSE Quadratic & 2variable inequalities practice questions with full step-by-step worked solutions. Covers difference of two squares, critical values, between the roots, outside the roots. Practise exam-style problems and check your method.

difference of two squarescritical valuesbetween the rootsoutside the rootsinclusive endpointsstrict endpoints

GCSE Higher70 questionsStep-by-step solutions

Question 1

2 markseasy

Solve

x^{2} < 9

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Worked solution

Find the critical values
$x^{2} - 9 = 0 \;\Rightarrow\; \left(x - 3\right) \left(x + 3\right) = 0 \;\Rightarrow\; x = -3 \quad \text{or} \quad x = 3$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 9$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-3$ and $3$ . We want the curve below the $x$ -axis, which is the part BETWEEN the critical values.
State the solution
$-3 < x < 3$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

-3 < x < 3

Question 2

1 markeasy

A dashed line is drawn with equation

y = 2x

. Which inequality describes the region above the dashed line (not including the line itself)?

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Worked solution

Write down the boundary line
$y = 2x$
The dashed line has equation $y = 2x$ .
Decide the inequality sign
$y > 2x \quad \text{(dashed, so strict)}$
The line is dashed, so points on the line are NOT included: the sign must be strict ( $>$ or $<$ ), never $\ge$ or $\le$ .
Decide above or below
$y > 2x$
Test $(1, 3)$ , which lies above the line: $3 > 2(1) = 2$ is true, so "above the line" matches $y > 2x$ . (A point below, such as $(2, 1)$ , gives $1 > 4$ , which is false.)

Answer

y > 2x

Question 3

2 marksintermediate

The region

R

satisfies

y < x + 2

y \ge -1

and

x \le 3

. Which statement about the three boundary lines of

R

is correct?

Show worked solution

Worked solution

Write down the inequalities that define $R$
$y < x + 2, \quad y \ge -1, \quad x \le 3$
A region in two variables is described by a list of inequalities; a point is in $R$ only if it satisfies every single one.
Deal with $y < x + 2$
$\text{boundary } y = x + 2 \quad \text{(dashed)}$
The boundary is the line $y = x + 2$ . The inequality excludes equality, so the line is drawn DASHED (points on it are NOT in R). The region wanted is below that line.
Deal with $y \ge -1$
$\text{boundary } y = -1 \quad \text{(solid)}$
The boundary is the line $y = -1$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is above that line.
Deal with $x \le 3$
$\text{boundary } x = 3 \quad \text{(solid)}$
The boundary is the line $x = 3$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is to the left of that line.
Sketch the region $R$
$y < x + 2, \quad y \ge -1, \quad x \le 3$
Draw every boundary line — solid where the inequality includes equality, dashed where it does not — then keep only the overlap of the required sides. $R$ is the piece of the plane satisfying ALL of the inequalities at once. (In the sketch, green lines are solid boundaries and red lines are dashed boundaries.)
State which boundaries are solid
$y < x + 2 \rightarrow \text{dashed}, \quad y \ge -1 \rightarrow \text{solid}, \quad x \le 3 \rightarrow \text{solid}$
Only the strict inequality gives a dashed line; the two inequalities with $\le$ or $\ge$ give solid lines.

Answer

y < x + 2 \rightarrow \text{dashed}, \quad y \ge -1 \rightarrow \text{solid}, \quad x \le 3 \rightarrow \text{solid}

Question 4

4 markshard

The region

R

satisfies

x \ge 0

y \ge 0

x + y \le 6

and

y \le 2x

. Work out the largest possible value of

x + 2y

for a point with integer coordinates in

R

Show worked solution

Worked solution

Write down the inequalities that define $R$
$x \ge 0, \quad y \ge 0, \quad x + y \le 6, \quad y \le 2 x$
A region in two variables is described by a list of inequalities; a point is in $R$ only if it satisfies every single one.
Deal with $x \ge 0$
$\text{boundary } x = 0 \quad \text{(solid)}$
The boundary is the line $x = 0$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is to the right of that line.
Deal with $y \ge 0$
$\text{boundary } y = 0 \quad \text{(solid)}$
The boundary is the line $y = 0$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is above that line.
Deal with $x + y \le 6$
$\text{boundary } y = 6 - x \quad \text{(solid)}$
The boundary is the line $y = 6 - x$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is below that line.
Deal with $y \le 2 x$
$\text{boundary } y = 2 x \quad \text{(solid)}$
The boundary is the line $y = 2 x$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is below that line.
Sketch the region $R$
$x \ge 0, \quad y \ge 0, \quad x + y \le 6, \quad y \le 2 x$
Draw every boundary line — solid where the inequality includes equality, dashed where it does not — then keep only the overlap of the required sides. $R$ is the piece of the plane satisfying ALL of the inequalities at once. (In the sketch, green lines are solid boundaries and red lines are dashed boundaries.)
Understand what is being maximised
$x + 2 y$
We only need points with WHOLE-NUMBER coordinates that lie in $R$ , and among those we want the biggest value of the expression. The best point is always in a corner of $R$ or as close to one as the integer grid allows.
Best point on the row $y = 0$
$y = 0: \quad x \le 6 \;\Rightarrow\; (6, 0), \quad (6) + 2 (0) = 6$
Fix $y = 0$ , push $x$ as far as the inequalities allow, and evaluate the expression.
Best point on the row $y = 1$
$y = 1: \quad x \le 5 \;\Rightarrow\; (5, 1), \quad (5) + 2 (1) = 7$
Fix $y = 1$ , push $x$ as far as the inequalities allow, and evaluate the expression.
State the maximum value
$x + 2 y = 10 \quad \text{at } (2, 4)$
The largest value of $x + 2 y$ at an integer point of $R$ is $10$ , reached at $(2, 4)$ .

Answer

10

Question 5

5 markschallenging

A region

R

is bounded by three straight lines.

R

lies strictly below the line

y = 2x + 3

, on or above the line

y = -x

, and strictly to the left of the line

x = 5

. Which set of inequalities defines

R

Show worked solution

Worked solution

Write down the inequalities that define $R$
$y < 2 x + 3, \quad y \ge - x, \quad x < 5$
A region in two variables is described by a list of inequalities; a point is in $R$ only if it satisfies every single one.
Deal with $y < 2 x + 3$
$\text{boundary } y = 2 x + 3 \quad \text{(dashed)}$
The boundary is the line $y = 2 x + 3$ . The inequality excludes equality, so the line is drawn DASHED (points on it are NOT in R). The region wanted is below that line.
Deal with $y \ge - x$
$\text{boundary } y = - x \quad \text{(solid)}$
The boundary is the line $y = - x$ . The inequality includes equality, so the line is drawn SOLID (points on it belong to R). The region wanted is above that line.
Deal with $x < 5$
$\text{boundary } x = 5 \quad \text{(dashed)}$
The boundary is the line $x = 5$ . The inequality excludes equality, so the line is drawn DASHED (points on it are NOT in R). The region wanted is to the left of that line.
Sketch the region $R$
$y < 2 x + 3, \quad y \ge - x, \quad x < 5$
Draw every boundary line — solid where the inequality includes equality, dashed where it does not — then keep only the overlap of the required sides. $R$ is the piece of the plane satisfying ALL of the inequalities at once. (In the sketch, green lines are solid boundaries and red lines are dashed boundaries.)
Translate "strictly below $y = 2x + 3$ "
$y < 2x + 3$
"Below" means $y$ is smaller than the value on the line; "strictly" means the line itself is excluded, so the sign is $<$ and the line is dashed.
Translate "on or above $y = -x$ "
$y \ge -x$
"Above" means $y$ is larger; "on or above" includes the line, so the sign is $\ge$ and the line is solid.
Translate "strictly to the left of $x = 5$ "
$x < 5$
To the left means smaller $x$ ; strictly means the vertical line $x = 5$ is dashed and excluded.
Rule out the option with $y > 2x + 3$
$y > 2x + 3 \;\text{is ABOVE the line}$
That option shades the wrong side of the first line.
Rule out the option with $y \le -x$
$y \le -x \;\text{is BELOW the line}$
That option shades the wrong side of the second line.
Rule out the option with $x > 5$
$x > 5 \;\text{is to the RIGHT}$
That option is on the wrong side of the vertical line.
Rule out the option with the wrong boundary types
$y \le 2x + 3, \; y > -x, \; x \le 5$
Here every boundary type is wrong: the two dashed lines have become solid and the solid line has become dashed.
Test the point $(1, 1)$
$(1, 1): \quad 1 < 5 \;\checkmark, \quad 1 \ge -1 \;\checkmark, \quad 1 < 5 \;\checkmark$
Substitute $x = 1$ and $y = 1$ into each inequality in turn. The point satisfies every inequality, so it lies in $R$ .
Test the point $(6, 1)$
$(6, 1): \quad 1 < 15 \;\checkmark, \quad 1 \ge -6 \;\checkmark, \quad 6 < 5 \;\times$
Substitute $x = 6$ and $y = 1$ into each inequality in turn. The point fails at least one inequality, so it does NOT lie in $R$ .
State the inequalities that define $R$
$y < 2x + 3, \quad y \ge -x, \quad x < 5$
Each phrase in the description turns into exactly one inequality, with the strictness matching solid/dashed.

Answer

y < 2x + 3, \quad y \ge -x, \quad x < 5

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