Simultaneous equationsEdexcel International A Level Maths: Revision notes

Simultaneous equations are two or more equations that must be true at the same time. The solution is a pair of values $(x,y)$ that satisfies every equation. For two linear equations you can use elimination or substitution. For substitution, rearrange one equation to make a variable the subject, then substitute it into the other. Example: $3x+2y=7$ and $x-y=4$. From the second, $x=y+4$. Then $3(y+4)+2y=7$, so $5y=-5$ and $y=-1$. Then $x=-1+4=3$. Check in both original equations: $3(3)+2(-1)=7$ and $3-(-1)=4$.

When one equation is linear and the other is non-linear, always substitute from the linear one. Rearrange the linear equation to make $y$ (or $x$) the subject. Example: $y=x+1$ and $x^2+y^2=25$.

1. Substitute: $x^2+(x+1)^2=25$.
2. Expand and rearrange to $=0$: $2x^2+2x-24=0$, so $x^2+x-12=0$.
3. Solve: $(x+4)(x-3)=0$, so $x=-4$ or $x=3$.
4. Back-substitute into the linear equation: $y=-3$ or $y=4$. The solutions are $(-4,-3)$ and $(3,4)$.

A quadratic gives two values of $x$, and each has its own $y$-value. State them as coordinate pairs, or as '$x=3,\ y=4$ or $x=-4,\ y=-3$'. Do not mix up the pairs. With $y=x+1$ the pair for $x=3$ is $y=4$, not $y=-3$. Always check each pair in both equations. For $(-3,-2)$: $y=x+1$ works, but $(-3)^2+(-2)^2=13\neq25$, so it is not a solution. A pair is only a solution if it satisfies both equations. If a question asks for the 'other solution', find the second pair after you have been given or found the first.

Each equation represents a line or curve. The solutions are the points of intersection. Example: $L$: $y=2x-5$ and $C$: $y=x^2-3x+1$. Equating gives $x^2-5x+6=0$, so $x=2$ or $x=3$, and the points are $(2,-1)$ and $(3,1)$. After substituting, the number of solutions of the resulting quadratic gives the number of intersections:

• two real roots: the line cuts the curve at two points;
• one repeated root: the line is a tangent;
• no real roots: the line and curve do not meet. Example: $y=2x-1$ and $y=x^2$ give $x^2-2x+1=(x-1)^2=0$, one solution $(1,1)$, so the line touches the curve.

If the quadratic does not factorise, use the quadratic formula or complete the square, and leave the answers in surd form when the question asks for exact values. Example: $y=x+2$ meets $y=x^2-3x+1$ where $x^2-4x-1=0$. Then $x=\frac{4\pm\sqrt{20}}{2}=2\pm\sqrt5$. Substitute into the linear equation: $y=4\pm\sqrt5$. The points are $(2+\sqrt5,\,4+\sqrt5)$ and $(2-\sqrt5,\,4-\sqrt5)$. Match the signs: the $+$ in the $x$-coordinate goes with the $+$ in the $y$-coordinate.

Define variables, write one equation for each piece of information, then solve by substitution. Example: a rectangle has perimeter $34$ cm and diagonal $13$ cm. Then $x+y=17$ and $x^2+y^2=169$. Substituting $y=17-x$ gives $x^2-17x+60=0$, so $x=12$ or $x=5$. With $x>y$ the sides are $12$ cm and $5$ cm. A diagonal of $12$ cm with the same perimeter gives $2x^2-34x+145=0$. Its discriminant is $1156-1160=-4<0$, so no such rectangle exists. Always check that the solutions suit the context, for example lengths must be positive.