Solving Inequalities: past paper questions & worked solutions

Divide by \(-4\), reversing the inequality: \[ x\ge-2 \]

R lies to the left of the \(y\)-axis, above \(y=-x\) and below \(2x+5y=12\).

\((-1,2)\): \(-1\le0\), \(2\ge1\), \(-2+10=8\le12\) ✓

\(x\le0\).

R lies to the left of the y-axis.

\(y\ge-x\).

R lies above the line y=-x.

\(2x+5y\le12\).

R lies below the other line. All three boundaries are solid, so equality is included.

Factorise:

The quadratic is greater than 0 outside the roots.

Factorise: \[ (2x+5)(2x-3)<0 \]

The quadratic is negative between the roots.

Factorise:

The quadratic is greater than 0 outside the roots.

Solve 2*x*² − 7*x* − 15 \> 0. Show clear algebraic working.

Factorise the quadratic: 2*x*² − 7*x* − 15 = (2*x*+3)(*x*−5) *Check:* (2*x*+3)(*x*−5) = 2*x*² − 10*x* + 3*x* − 15 = 2*x*² − 7*x* − 15

Roots: set each factor to zero: $$2x+3=0 \\implies x=-\\tfrac{3}{2};\\qquad x-5=0 \\implies x=5$$

Since the coefficient of *x*² is  + 2 \> 0, the parabola opens **upward**. The expression is  \> 0 *outside* the roots:

Factorise:

The quadratic is less than 0 between the roots.

Since \(n\) is an integer, \[ n=-2,-1,0,1,2 \]

(b) For \[ y\le1 \]

use a closed circle at \(1\) and shade to the left.

The region is to the right of the \(y\)-axis: \[ x\geq0 \]

It is above the line \(y=x\): \[ y\geq x \]

The bottom boundary is \[ y=-3 \] so \[ y\geq-3 \]

The rising boundary is \[ y=2x+2 \] so the region has \[ y\leq2x+2 \]

The falling boundary goes through \((0,1)\) and \((1,0)\), so \[ x+y=1 \]

The region is below this line: \[ x+y\leq1 \]

The shaded region is below this line: \[ y\leq3x-2 \]

The vertical boundary is \[ x=3 \] so the region has \[ x\leq3 \]

The lower boundary is the \(x\)-axis, so \[ y\geq0 \]

The shaded region is left of the \(y\)-axis: \[ x\leq0 \]

It is below the line through \((0,4)\) and \((4,0)\): \[ y\leq -x+4 \]

So \[ y\geq \frac13x-2 \]

Factorise:

So

The quadratic opens upwards, so it is less than or equal to zero between the roots.

For part (a), \(w<1\) is shown with an open circle at 1 and an arrow to the left.

For part (b), the region is right of \(x=-3\), below \(y=4\), and above \(y=x\): \[ x\geq-3 \]

Draw an open circle at 1 and an arrow pointing to the left.

**Step 1: Identify the vertical boundary.** The region is to the right of the vertical line at *x* =  − 3. Inequality 1: *x* ≥  − 3

**Step 2: Identify the horizontal boundary.** The region is below the horizontal line at *y* = 4. Inequality 2: *y* ≤ 4

**Step 3: Identify the diagonal boundary.** The line passes through (0,0), (1,1), (−1,−1), etc., meaning its equation is *y* = *x*. The shaded region is above this line. Inequality 3: *y* ≥ *x*

(a) The number line has an open circle at \(-2\), a closed circle at \(1\), and shading between them.

Draw the boundary lines: \[ x=3, \qquad y=1, \qquad x+y=7 \]

Draw the boundary lines: \[ y=2, \qquad x=6, \qquad y=x+1 \]

Factorise:

Since the coefficient of \(x^2\) is positive, the quadratic is greater than 0 outside the roots.

The vertical boundary is \[ x=1 \] so the region has \[ x\leq1 \]

The horizontal boundary is \[ y=-2 \] so \[ y\geq-2 \]

The sloping line passes through \((0,4)\) and has gradient \(2\), so \[ y=2x+4 \]

The shaded region is below this line: \[ y\leq2x+4 \]

Draw the boundary lines: \[ y=1, \qquad x=2, \qquad x+y=7 \]

Divide by 2: \[ -2\le y<3 \]

Since \(y\) is an integer, \[ y=-2,-1,0,1,2 \]

(a) For \[ -2\le y<1 \]

use a closed circle at \(-2\), an open circle at \(1\), and shade the line between them.

Draw the three boundary lines: \[ x=1.5, \qquad y=x, \qquad x+y=6 \]

For part (a), \[ 2<x\leq4 \]

For the region \(R\), the boundaries are \[ y=x-3, \qquad x+y=5, \qquad y=6 \]

The shaded region is above the two sloping lines and below the horizontal line: \[ y\geq x-3 \]

Draw the three boundary lines:

For \(x+y=7\), use points such as \((0,7)\) and \((7,0)\).

The required region is:

So shade the triangular region above \(y=1\), to the right of \(x=2\), and below the line \(x+y=7\). Its vertices are

Place an open circle at \(1\) and draw a horizontal ray from it to the left, with a left-pointing arrow. Do not fill the circle.

The strict inequality excludes \(1\) and includes every real number less than \(1\).

Subtract 3 from all three expressions.

The integer values in this interval are \(-1,0,1,2,3\).

Include the lower endpoint but exclude 4.

Subtract 8 from both sides.

Division by a negative number reverses the inequality sign.

For part (a), \[ 7-3x<2x+15 \]

For part (b), the boundaries are \(x=2\), \(y=3\), and \(x+y=9\).

(b) The three boundary lines are \(x=2\), \(y=3\), and \(x+y=9\). The shaded side gives

For part (a), \[ 8x-4\geq3x-10 \]

For part (b), the region is above \(y=2\), left of \(x=7\), and below \(y=x\): \[ y\geq2 \]

For part (a), \[ 9-4x>17 \]

For part (b), the region is above \(y=2\), left of \(x=6\), and below \(y=x\): \[ y\geq2 \]

(b)(ii) \(y=7\) or \(y=-5\).

A straight line with gradient \(-3\) passing through \((0,5)\) has equation \[ y=-3x+5 \]

For the region, use the inequalities: \[ x<6 \]

and

For the side lengths to be possible,

The area is less than \(75\):

Factorise:

So

Combining with \(x>1\):

Add 3 to each part: \[ -4\le2x<8 \]

Divide by 2: \[ -2\le x<4 \]

On the number line, use a closed circle at \(-2\), an open circle at \(4\), and shade between them.

Factorise: \[ (x-8)(x+5)=0 \]

So \[ x=8 \quad \text{or} \quad x=-5 \]

\(a\) Solve 3*x* − 22 \> 8*x* − 40 algebraically. (b) Write the three inequalities defining region R.

Move all *x* terms to the right and constants to the left: 3*x* − 22 \> 8*x* − 40 ⟹  − 22 + 40 \> 8*x* − 3*x* ⟹ 18 \> 5*x*

Divide both sides by 5: $$x \< \\frac{18}{5} = 3.6$$

*x* \< 3.6

Identify the three boundary lines and their regions using the graph. The shaded region R is bounded by *x* = 2, *y* = 1, and *x* + *y* = 6.

(a)(ii) \(x=-8\) or \(x=3\).

Multiply by 10: \[ 2(4-3x)-5(3x-5)=-30 \]

Factorise: \[ (5y+8)(y-5)\le0 \]

The roots are \(-\frac85\) and \(5\). The quadratic is \(\le0\) between the roots.

Multiply 4e by both terms in the bracket.

Multiply each term in the first bracket by each term in the second and collect like terms.

From \(t=ax-2n\), add \(2n\) to give \(t+2n=ax\).

Use the same operation on both sides.

Divide by the coefficient of x.

Draw an open circle at \(y=-1\), a filled circle at \(y=3\), and a straight segment joining them. The represented interval is \((-1,3]\).

The strict lower inequality excludes -1; the non-strict upper inequality includes 3.

Multiply by 6: \[ 2(4x+5)-3(3-2x)=78 \]

Factorise: \[ (2y+5)(y-6)\le0 \]

The roots are \(-\frac52\) and \(6\). The quadratic is \(\le0\) between the roots.

(b) \[ \frac{y^5\times y^n}{y^6}=y^{n-1}=y^{13} \] so \(n-1=13\), hence \(n=14\).

On the number line, use an open circle at 3 and shade to the left.

(a) \(g^9\div g^2=g^7\).

(b) \(5k^2(k^3+4)=5k^5+20k^2\).

(c)(ii) \(x=9\) or \(x=-7\).