Hard GCSE Quadratic & 2variable inequalities Questions

Write down the inequalities that define $R$

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 $y < 2 x + 3$

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 $y \ge - x$

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 $x < 5$

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 $R$

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 $y = 2x + 3$"

"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 $y = -x$"

"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 $x = 5$"

To the left means smaller $x$; strictly means the vertical line $x = 5$ is dashed and excluded.

Rule out the option with $y > 2x + 3$

That option shades the wrong side of the first line.

Rule out the option with $y \le -x$

That option shades the wrong side of the second line.

Rule out the option with $x > 5$

That option is on the wrong side of the vertical line.

Rule out the option with the wrong boundary types

Here every boundary type is wrong: the two dashed lines have become solid and the solid line has become dashed.

Test the point $(1, 1)$

Substitute $x = 1$ and $y = 1$ into each inequality in turn. The point satisfies every inequality, so it lies in $R$.

Test the point $(6, 1)$

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 $R$

Each phrase in the description turns into exactly one inequality, with the strictness matching solid/dashed.

Write down the inequality

The quadratic is already compared with zero, so we can go straight to the critical values.

Multiply by $-1$ and reverse the inequality sign

Multiplying (or dividing) an inequality by a negative number reverses the sign. Doing this makes the $x^2$ coefficient positive, so the parabola is the usual U-shape.

Factorise the quadratic

Factorising gives the two brackets whose zeros are the critical values.

Find the critical values

The critical values are where the curve cuts the $x$-axis. They split the number line into three parts.

Sketch the parabola

The coefficient of $x^2$ is positive, so the curve is a U-shape crossing the $x$-axis at $-2$ and $3$. The sketch is what tells you WHICH side you need — never guess.

Read off the part of the curve below the $x$-axis

We want $x^{2} - x - 6$ to be negative or zero, i.e. the part of the U-shape that dips BELOW the $x$-axis. That is the region BETWEEN the two critical values.

Check a value inside the solution set ($x = \frac{1}{2}$)

Substituting a value from inside the claimed solution set into the ORIGINAL inequality makes it true, as it must.

Check a value outside the solution set ($x = -3$)

A value from outside the claimed solution set makes the original inequality false — this is the check that catches a solution written on the wrong side of the critical values.

Test the critical value $x = -2$

At a critical value the quadratic equals $0$. The inequality is not strict ($\,\le\,$), so this endpoint is included.

Test the critical value $x = 3$

Same story at the other critical value: the quadratic is $0$ there, so this endpoint is included too.

Write the solution in set notation

Set notation reads "the set of values of $x$ such that ...". A single interval needs only one set.

Write the solution in interval notation

Round brackets exclude an endpoint, square brackets include it — so they must match the strict/non-strict signs exactly.

Check a value on the other side ($x = 4$)

The original inequality is false here too, confirming that only the middle strip works.

Locate the minimum point of the parabola

The minimum sits midway between the roots, and its $y$-value is negative — which is exactly why the curve dips below the axis between the critical values.

Explain the error and give the correct solution

The critical values were right; the mistake was choosing the wrong side of the parabola. The sketch decides it — never the order in which the numbers appear.

Write down the inequalities that define $R$

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 $x \ge 0$

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 $y \ge 1$

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 $y \le 2 x + 1$

The boundary is the line $y = 2 x + 1$. 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 $x + 2 y \le 10$

The boundary is the line $y = 5 - \frac{x}{2}$. 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 $R$

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

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 $y = 1$

Fix $y = 1$, push $x$ as far as the inequalities allow, and evaluate the expression.

Best point on the row $y = 2$

Fix $y = 2$, push $x$ as far as the inequalities allow, and evaluate the expression.

Best point on the row $y = 3$

Fix $y = 3$, push $x$ as far as the inequalities allow, and evaluate the expression.

Best point on the row $y = 4$

Fix $y = 4$, push $x$ as far as the inequalities allow, and evaluate the expression.

Test the point $(2, 4)$

Substitute $x = 2$ and $y = 4$ into each inequality in turn. The point satisfies every inequality, so it lies in $R$.

Compare the rows

The largest of the row maxima is the overall maximum.

Why not go one step further? Test $(2, 5)$

Pushing $y$ up by one from the best point would give a bigger value of $x + 3 y$, but the point leaves $R$ — so it is not allowed.

State the maximum value

The largest value of $x + 3 y$ at an integer point of $R$ is $14$, reached at $(2, 4)$.

Write down the inequalities that define $R$

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 $y \ge 0$

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 $y \le x$

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 below that line.

Deal with $x + y \le 8$

The boundary is the line $y = 8 - x$. 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 $R$

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.)

Count the integer points with $x = 0$

With $x = 0$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 1$

With $x = 1$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 2$

With $x = 2$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 3$

With $x = 3$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 4$

With $x = 4$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 5$

With $x = 5$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 6$

With $x = 6$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 7$

With $x = 7$ fixed, the inequalities pin $y$ down to these whole numbers.

Count the integer points with $x = 8$

With $x = 8$ fixed, the inequalities pin $y$ down to these whole numbers.

State the number of integer points

There are $25$ points with whole-number coordinates inside or on the included boundaries of $R$.

Write down the two inequalities

The point must satisfy BOTH inequalities to be inside the region.

Substitute the point $(k, 2)$

Every point of the region has to obey both inequalities, and this point has $y = 2$.

Substitute into the first inequality

Replace $y$ with $2$ and $x$ with $k$ in $y < 2x - 1$.

Solve the first inequality

Add $1$ to both sides, then divide by the positive number $2$: the sign does not flip.

Substitute into the second inequality

Replace $x$ with $k$ and $y$ with $2$ in $x + y < 7$.

Solve the second inequality

Subtract $2$ from both sides.

Both conditions must hold

The point is inside the region only if BOTH inequalities are satisfied, so we take the overlap of the two solution sets.

Write as a single inequality

Both signs are strict, because both boundary lines are dashed (the inequalities are $<$, not $\le$).

Check $k = 3$

The point $(3, 2)$ satisfies both inequalities, so it lies inside the region.

Check $k = \frac{3}{2}$

At $k = \tfrac{3}{2}$ the point sits exactly ON the dashed line $y = 2x - 1$, and $2 < 2$ is false — so this endpoint is excluded.

Check $k = 5$

At $k = 5$ the point sits exactly ON the dashed line $x + y = 7$, so it is excluded too.

Check $k = 1$

$k = 1$ is below $\tfrac{3}{2}$, and it fails the first inequality, as expected.

Check $k = 6$

$k = 6$ passes the first inequality but fails the second — a reminder that BOTH must hold.

Write the answer in set notation

The set of values of $k$ for which $(k, 2)$ lies inside the region.

State the range of $k$

Any $k$ strictly between $\tfrac{3}{2}$ and $5$ puts the point $(k, 2)$ inside the region; the endpoints themselves land on the dashed boundaries and are excluded.