Substitution turns a linear/quadratic pair into a single quadratic; the discriminant of that quadratic then tells you how many intersection points the line and curve have. Inequalities are best solved by sketching the sign of a factorised expression rather than by manipulating symbols.
Simultaneous equations (MM1.4)
- Make one variable the subject of the linear equation and substitute into the quadratic one.
- The number of solutions equals the number of intersection points of the line and the curve; use the discriminant of the resulting quadratic.
- Symmetric tricks: from \(x + y\) and \(xy\) (or \(x^2 + y^2\)) you can find \((x - y)^2 = (x + y)^2 - 4xy\) without solving.
- \(x^2 = y^2\) means \(y = x\) or \(y = -x\): two cases.
Inequalities (MM1.5, M4.17)
- Linear: as equations, but multiplying or dividing by a negative reverses the sign.
- Quadratic: factorise, find the roots, sketch, read off where the graph is above or below the axis. \((x - 2)(x - 5) < 0\) gives \(2 < x < 5\); \(\geq 0\) gives the two outer regions.
- Products of three or more factors: use a sign chart (the sign changes at each simple root).
- Never multiply an inequality by an expression whose sign is unknown (\(x\), \(x - 2\)); either split into cases or bring everything to one side and study the sign of a fraction.
- Modulus: \(|x - a| < r\) means \(a - r < x < a + r\); \(|x| > r\) means \(x < -r\) or \(x > r\).
- 'Both inequalities': intersect the solution sets on a number line; 'either': take the union.
Must-know (no formula booklet)
- \((x - p)(x - q) < 0 \Leftrightarrow p < x < q\) (for \(p < q\))
- \(|x - a| < r \Leftrightarrow a - r < x < a + r\)
- one line-curve intersection \(\Leftrightarrow\) discriminant \(= 0\)
Classic traps
- \(\frac1x > 2\) is not \(1 > 2x\): for \(x < 0\) the inequality flips. Solution: \(0 < x < \frac12\).
- Replacing \(x\) by \(-x\) in \(ax^2 + bx + c < 0\) reflects the solution set, so \(-2 < x < 5\) becomes \(-5 < x < 2\).
- Count integers in an interval carefully: \(2 < n < 8\) contains five integers.
Worked example
Q. Solve \(x + 2y = 7\), \(x^2 + y^2 = 10\).
A. \(x = 7 - 2y\): \(5y^2 - 28y + 39 = 0\), \((5y - 13)(y - 3) = 0\), giving \((1, 3)\) and \((\frac95, \frac{13}{5})\).