Questions on simultaneous equations and inequalities

Question 1

Solve the simultaneous equations 2x + y = 7 and x − y = 2.

Accepted Answer

Adding the two equations eliminates y, giving 3x = 9, so x = 3, and substituting back into either equation gives y = 1. M1 for a correct elimination or substitution, A1 for x, A1 for y. Putting (3, 1) into both original equations is a one-line check on the arithmetic.

Question 2

Solve the inequality x² − x − 12 < 0.

Accepted Answer

Factorise to (x − 4)(x + 3) < 0, so the boundary values are x = −3 and x = 4. The parabola opens upwards, so it lies below the axis strictly between its roots, giving −3 < x < 4. M1 for factorising, A1 for the roots, A1 for the inequality. Quote the roots, then read the sign off the shape, and never simply write the roots down joined by inequality signs without checking which way round they go.

Question 3

Solve the inequality 4 − 3x > 13.

Accepted Answer

Subtract 4: −3x > 9. Dividing by −3 reverses the inequality, so x < −3. M1 for isolating the x term, A1 for the reversed inequality. A test value settles the direction: x = −4 gives 4 + 12 = 16 > 13, true, while x = 0 gives 4 > 13, false, so the solutions lie on the negative side. Dividing by a negative number without reversing gives x > −3, which contains x = 0 and is wrong at the first value tried. Adding 3x to both sides instead, 4 > 13 + 3x, avoids the negative division altogether and gives −9 > 3x, the same answer.

Question 4

On a sketch, shade the region of the plane whose points satisfy both y > x + 1 and x + y ≤ 3.

Accepted Answer

Draw the boundary y = x + 1 as a dotted line, because the inequality is strict and points on the line are excluded. B1. Draw x + y = 3, that is y = 3 − x, as a solid line, because ≤ includes its boundary. B1. The lines cross at (1, 2). Test a point: (0, 2) gives 2 > 1 and 0 + 2 ≤ 3, both true, so the region is on that side of each line, above the dotted line and below the solid one, the wedge opening to the left from (1, 2). B1 for the correct region shaded. The convention carries meaning: a dotted boundary says the edge is not part of the answer, so a point such as (0, 1) on y = x + 1 fails the first inequality even though it lies on the drawn line, while (0, 3) on the solid line satisfies both.

Question 5

Find the coordinates of the points of intersection of the line y = x + 1 and the curve y = x² − 3x + 4.

Accepted Answer

Equating the two expressions gives x² − 3x + 4 = x + 1, so x² − 4x + 3 = 0, which factorises as (x − 1)(x − 3) = 0. Then x = 1 gives y = 2 and x = 3 gives y = 4, so the points are (1, 2) and (3, 4). M1 for equating, A1 for the three-term quadratic, M1 for solving, A1 for the x values, A1 for the y values. The answers come in matched pairs, so each x has to be given with the y found from the same equation.

Question 6

The line y = 2x + c is a tangent to the curve y = x². Find the value of c.

Accepted Answer

Substituting gives x² = 2x + c, so x² − 2x − c = 0. Tangency means the line meets the curve exactly once, so this quadratic has a repeated root and its discriminant is zero: b² − 4ac = 4 + 4c = 0, giving c = −1. M1 for substituting, A1 for the quadratic, M1 for setting the discriminant to zero, A1 for c. Positive means two crossings, zero means a tangent, negative means the line misses the curve entirely.

Question 7

Solve the inequality 2/x > 1.

Accepted Answer

Multiplying by x is unsafe, because the inequality flips whenever x is negative. Multiply by x² instead, which is positive for every x ≠ 0: 2x > x², so x² − 2x < 0 and x(x − 2) < 0, giving 0 < x < 2. M1 for a valid method that handles the sign, A1 for the quadratic inequality, A1 for the interval, B1 for checking. Test three values: x = 1 gives 2 > 1, true; x = 3 gives 2/3 > 1, false; x = −1 gives −2 > 1, false. Multiplying naively by x produces x < 2 and wrongly hands the whole negative axis to the solution set.

Question 8

Find the set of values of x for which x² + 2x − 15 ≥ 0, giving your answer in set notation.

Accepted Answer

Factorise to (x + 5)(x − 3) ≥ 0, with boundary values −5 and 3. The parabola opens upwards, so it lies on or above the axis outside its roots, giving x ≤ −5 or x ≥ 3. In set notation that is {x : x ≤ −5} ∪ {x : x ≥ 3}. M1 for factorising, A1 for the two regions, A1 for correct set notation. The two pieces need the union symbol rather than a comma, since they are separate stretches of the number line with a gap between them, and writing 3 ≤ x ≤ −5 describes nothing at all.

Question 9

The region R consists of the points satisfying y < 4 − x² and y ≥ x + 2. Find the x-coordinates of the points where the two boundary curves meet, and shade R on a sketch, using the correct convention for each boundary.

Accepted Answer

The boundaries meet where x + 2 = 4 − x², so x² + x − 2 = 0 and (x + 2)(x − 1) = 0. M1 for equating and forming the quadratic, A1 for x = −2 and x = 1, the points (−2, 0) and (1, 3). The parabola y = 4 − x² is drawn dotted, since y < excludes it, and the line y = x + 2 is drawn solid, since y ≥ includes it. B1 for the two conventions. R is below the parabola and on or above the line: the part of the plane between the two meeting points, bounded above by the arch of the parabola and below by the line. Test (0, 3): 3 < 4 and 3 ≥ 2, so it lies inside. B1 for the region. The parabola opens downwards with vertex (0, 4), which is where the arch is; drawing it upwards puts the region on the outside of the curve and unbounded. The meeting points (−2, 0) and (1, 3) lie on the dotted boundary as well as the solid one, so they are not in R.

Question 10

Solve the simultaneous equations x − y = 1 and xy = 12.

Accepted Answer

Make x the subject of the linear equation, x = y + 1, and substitute into the product: (y + 1)y = 12, so y² + y − 12 = 0. M1 for substituting, A1 for the quadratic. It factorises as (y + 4)(y − 3) = 0, so y = 3 or y = −4. A1 for both y values. Then x = y + 1 gives x = 4 and x = −3: the solutions are x = 4, y = 3 and x = −3, y = −4. A1 for both pairs, each x with its own y. Checking, 4 × 3 = 12 and (−3)(−4) = 12, and both pairs differ by 1. The pairs are not interchangeable: x = 4 with y = −4 has a product of −16. Two solutions were to be expected, since the line x − y = 1 cuts both branches of the reciprocal curve xy = 12.

Question 11

Solve the inequality (x − 2)(x + 5) > x(x − 1).

Accepted Answer

Expand both sides: x² + 3x − 10 > x² − x. M1 for expanding. The x² terms cancel, leaving 3x − 10 > −x, so 4x > 10. M1 for collecting. Hence x > 5/2. A1. The brackets suggest a quadratic inequality, but the squared terms are equal on both sides and cancel, so the boundary is a single value rather than two and no sketch of a parabola is needed. Setting each bracket on the left to zero, x = 2 and x = −5, treats the question as (x − 2)(x + 5) > 0, which is a different inequality. A check at x = 3: the left is 8 and the right is 6, so 3 belongs, and at x = 2 the left is 0 and the right is 2, so 2 does not.

Question 12

Solve the simultaneous equations x² + y² = 25 and y = 2x − 5.

Accepted Answer

Substitute the linear equation into the quadratic one: x² + (2x − 5)² = 25, so x² + 4x² − 20x + 25 = 25, which reduces to 5x² − 20x = 0. Factorising, 5x(x − 4) = 0, so x = 0 or x = 4. Then y = 2(0) − 5 = −5 and y = 2(4) − 5 = 3, giving the points (0, −5) and (4, 3). M1 for substituting, A1 for the quadratic, M1 for solving, A1 for the x values, A1 for the y values. Substitute the line into the circle rather than the other way round, since making y the subject of x² + y² = 25 drags a square root through the whole calculation for no benefit.

Question 13

Find the set of values of k for which the line y = kx − 3 does not intersect the curve y = x² + x + 1.

Accepted Answer

Setting the two equal gives x² + x + 1 = kx − 3, so x² + (1 − k)x + 4 = 0. No intersection means this quadratic has no real roots, so its discriminant is negative: (1 − k)² − 16 < 0, that is (1 − k)² < 16. Taking square roots carefully, −4 < 1 − k < 4, so −3 < k < 5. M1 for equating, A1 for the quadratic in x, M1 for the discriminant condition, A1 for (1 − k)² < 16, M1 for solving, A1 for the interval. Two traps sit here. The discriminant involves the coefficients of the substituted quadratic, not of the original curve, and (1 − k)² < 16 gives an interval rather than 1 − k < 4 alone. Check with k = 0: the line y = −3 and the curve whose minimum is 0.75 clearly never meet, and 0 does lie in the answer.

Question 14

The point P has coordinates (0, 4) and the curve C has equation y = x² + 2x + 5. Show that there are exactly two lines through P that are tangents to C, find their equations, and find the coordinates of the point where each touches C.

Accepted Answer

A line through P with gradient m is y = mx + 4. Where it meets C, x² + 2x + 5 = mx + 4, so x² + (2 − m)x + 1 = 0. M1 for substituting the family of lines, A1 for the quadratic. A tangent meets the curve at a repeated root, so the discriminant is zero: (2 − m)² − 4 = 0, giving 2 − m = ±2, so m = 0 or m = 4. M1 for setting the discriminant to zero, A1 for both gradients, which shows there are exactly two tangents from P, one for each root of the quadratic in m. The lines are y = 4 and y = 4x + 4. For the points of contact, the repeated root of the substituted quadratic gives the x-coordinate. With m = 0 it is x² + 2x + 1 = (x + 1)², so x = −1 and y = 4: the point (−1, 4). With m = 4 it is x² − 2x + 1 = (x − 1)², so x = 1 and y = 8: the point (1, 8). M1 for using the repeated root, A1 for both points. The first tangent is horizontal, and (−1, 4) is the vertex of the parabola, since x² + 2x + 5 = (x + 1)² + 4; P sits below the curve, at height 4 when the curve at x = 0 is at 5, which is why two tangents exist. A vertical line through P is not a tangent and is not of the form y = mx + 4, so the family misses nothing.

Question 15

A rectangular garden has a perimeter of 34 m and a diagonal of length 13 m. By forming and solving a pair of simultaneous equations, find the dimensions of the garden.

Accepted Answer

Let the sides be x m and y m. The perimeter gives 2x + 2y = 34, so x + y = 17, and the diagonal gives, by Pythagoras, x² + y² = 169. M1 for forming both equations from the context, A1 for the pair. Substitute y = 17 − x into the second: x² + (17 − x)² = 169, so 2x² − 34x + 289 = 169 and 2x² − 34x + 120 = 0, or x² − 17x + 60 = 0. M1 for the substitution, A1 for the quadratic. It factorises as (x − 5)(x − 12) = 0, so x = 5 with y = 12, or x = 12 with y = 5, the same rectangle either way: the garden is 5 m by 12 m. A1. Checking, 2(5 + 12) = 34 and 5² + 12² = 25 + 144 = 169. Writing the perimeter as x + y = 34 gives x² − 34x + 493.5 = 0, whose discriminant is negative, and the model then has no rectangle at all: the factor 2 belongs in the first equation.