Solving linear and non-linear simultaneous equations

A pair of simultaneous equations in which one equation is linear and the other contains \(x^2\), \(y^2\) or \(xy\) is solved by substitution. Each solution is a pair \((x,\,y)\); graphically, the pairs are the points where the line meets the curve.

Why: the linear equation lets you write one unknown in terms of the other. Substituting into the non-linear equation leaves one equation in one unknown, usually a quadratic, which can have two, one or no real roots.

Method

1. Make \(x\) or \(y\) the subject of the linear equation, choosing the one that avoids fractions.
2. Substitute into the non-linear equation and simplify to a quadratic equal to zero.
3. Solve the quadratic (factorise, or use the formula and give the accuracy asked for).
4. Substitute each value back into the linear equation to find its partner.
5. Write the answers as pairs, or as coordinates when the question asks for points of intersection, and check one pair in both equations.

Example

Solve \(y=2x-1\) and \(x^2+xy=2\).

\(x^2+x(2x-1)=2\), so \(3x^2-x-2=0\) and \((3x+2)(x-1)=0\).

\(x=1\): \(y=1\). \(x=-\dfrac23\): \(y=2\left(-\dfrac23\right)-1=-\dfrac73\).

\(\therefore\) \(x=1,\ y=1\) or \(x=-\dfrac23,\ y=-\dfrac73\). Check: \(1+1(1)=2\).

Setting up from a description

Name the unknowns, then write one equation for each fact. A rectangle with sides \(x\) cm and \(y\) cm has perimeter \(14\) cm and diagonal \(5\) cm: \(x+y=7\) and \(x^2+y^2=25\). Substituting \(y=7-x\) gives \(x^2-7x+12=0\), so the sides are \(3\) cm and \(4\) cm.