Simultaneous equationsAQA A-Level Maths: Revision notes

Simultaneous equations are equations that must be true at the same time. Solving them finds the values of the unknowns that satisfy every equation. Graphically, each solution is a point where the graphs intersect. Two linear equations in two unknowns usually have exactly one solution. A line and a quadratic curve can have two solutions, one (the line is a tangent) or none. There are two methods: elimination and substitution. Always check your answer in both original equations.

Multiply one or both equations so that the coefficients of one unknown match, then add or subtract to eliminate it. $x+2y=7,\qquad 3x-y=7.$ Multiply the second by 2: $6x-2y=14$. Adding gives $7x=21$, so $x=3$, and then $y=2$. Check: $3+4=7$ and $9-2=7$. Subtract when the coefficients have the same sign and add when they have opposite signs.

Rearrange one equation to make a variable the subject and substitute it into the other. This is the method for a linear and a quadratic equation, because elimination cannot remove a squared term. Worked example. $y=x+1$ and $y=x^2-5$. Substituting: $x+1=x^2-5$, so $x^2-x-6=0$ and $(x-3)(x+2)=0$. Then $x=3$ gives $y=4$ and $x=-2$ gives $y=-1$. Always substitute back into the linear equation to find the matching $y$, and give the solutions as pairs: $(3,4)$ and $(-2,-1)$.

Substitute the linear equation into the quadratic one. Expand carefully and collect to form $ax^2+bx+c=0$, which gives up to two values of $x$. Example: $2x+y=k$ and $x^2+y^2=20$. Using $y=k-2x$: $x^2+(k-2x)^2=20$, so $5x^2-4kx+k^2-20=0$. Expand $(k-2x)^2$ as $k^2-4kx+4x^2$ with all three terms. Then solve by factorising or the formula.

After substitution, the discriminant of the resulting quadratic tells you how the line meets the curve:

• $b^2-4ac>0$: the line meets the curve at two points.
• $b^2-4ac=0$: the line is a tangent (one point, a repeated root).
• $b^2-4ac<0$: the line and curve do not meet. For $5x^2-4kx+k^2-20=0$: $\Delta=16k^2-20(k^2-20)=400-4k^2$, which is zero when $k=\pm10$. The same idea proves that a situation is impossible, such as a rectangle of perimeter 34 cm with diagonal 12 cm: $2x^2-34x+145=0$ has $\Delta=-4<0$.

Define the unknowns, translate each fact into an equation, then solve. A rectangle with perimeter 34 cm and diagonal 13 cm gives $x+y=17$ and $x^2+y^2=169$. Substituting $y=17-x$ gives $x^2-17x+60=0$, so $x=5$ or $12$. The pair of sides is 5 cm and 12 cm. Check the answers make sense in context: lengths must be positive, and each solution pair must satisfy both equations.