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Simultaneous equations and inequalities questions
Two equations sharing their unknowns are a picture of a line crossing a curve, and the algebra lands on the crossings. Those same pictures then answer a further question. An inequality identifies the values for which one expression is greater or less than the other, and the answer becomes a region of the plane.
8 original questions · 33 marks · the simultaneous equations and inequalities notes · Algebra and functions
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Solve the simultaneous equations 2x + y = 7 and x − y = 2.
Worked 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 the one-line check the answer deserves, and it costs less time than losing the accuracy marks.Solve the inequality x2 − x − 12 < 0.
Worked 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. A sketch is the argument here. 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.Find the coordinates of the points of intersection of the line y = x + 1 and the curve y = x2 − 3x + 4.
Worked 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, and pairing them the wrong way round costs the final mark even when both lists are right.The line y = 2x + c is a tangent to the curve y = x2. Find the value of c.
Worked 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. The discriminant of the substituted quadratic is geometry in disguise. Positive means two crossings, zero means a tangent, negative means the line misses the curve entirely.Solve the inequality 2/x > 1.
Worked 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.Find the set of values of x for which x2 + 2x − 15 ≥ 0, giving your answer in set notation.
Worked 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.Solve the simultaneous equations x2 + y2 = 25 and y = 2x − 5.
Worked 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.Find the set of values of k for which the line y = kx − 3 does not intersect the curve y = x2 + x + 1.
Worked 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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise simultaneous equations and inequalities one question at a time
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