Solving simultaneous equations: two linear equations by elimination and by substitution, finding the solution graphically, and solving one linear and one quadratic equation (Higher tier).

What this dot point is asking

Simultaneous equations are two equations that share the same two unknowns; solving them means finding the values that satisfy both at once. Edexcel expects you to solve two linear equations by elimination and by substitution, to recognise the solution as the intersection of two graphs, and at Higher tier to solve one linear with one quadratic equation. Geometrically, you are finding where two lines (or a line and a curve) cross.

The elimination method

Elimination removes one variable by adding or subtracting the equations. The rule of thumb: if the matching coefficients have the same sign, subtract; if opposite signs, add.

For $2x + 3y = 13$ and $4x - 3y = 5$, the $y$ coefficients are $+3$ and $-3$, so add: $6x = 18$, giving $x = 3$. Substitute back to find $y$: $2(3) + 3y = 13$, so $y = \tfrac{7}{3}$. When neither variable matches, scale first: to solve $3x + 2y = 16$ and $2x + 5y = 18$, multiply the first by $5$ and the second by $2$ to make the $y$ coefficients both $10$, then subtract.

The substitution method

Substitution is often easiest when one equation already gives a variable in terms of the other.

The graphical method

Each linear equation is a straight line. Plotting both lines, their point of intersection gives the simultaneous solution, because that point lies on both lines. If the lines are parallel they never meet, so there is no solution; if they are the same line, there are infinitely many. The graphical method is approximate unless the intersection falls on exact grid points, so it is usually a way to check rather than to find exact answers.

One linear and one quadratic (Higher)

When one equation is quadratic and the other linear, substitution is the method. Rearrange the linear equation for one variable and substitute into the quadratic, producing a single quadratic equation in one unknown. Solve it (by factorising or the formula), then substitute each solution back into the linear equation to find the paired value. Because a line can cut a curve in two places, expect up to two solution pairs. Geometrically, you are finding where a straight line meets a parabola or circle.

A frequent Higher version pairs a line with a circle, such as $x^2 + y^2 = 25$ and $y = x + 1$. Substitute $y = x + 1$ into the circle: $x^2 + (x+1)^2 = 25$, which expands to $2x^2 + 2x - 24 = 0$, or $x^2 + x - 12 = 0$ after dividing by $2$. This factorises as $(x + 4)(x - 3) = 0$, so $x = -4$ or $x = 3$, giving the two crossing points $(-4, -3)$ and $(3, 4)$. The method is identical to the parabola case; only the quadratic that appears is different.

Try this

Q1. Solve $x + y = 10$ and $x - y = 4$. [3 marks]

• Cue. Add the equations: $2x = 14$, so $x = 7$; then $y = 3$.

Q2. Solve $y = x + 2$ and $y = x^2$. [3 marks]

• Cue. Set $x^2 = x + 2$, so $x^2 - x - 2 = 0$, giving $(x-2)(x+1) = 0$. Then $x = 2, y = 4$ or $x = -1, y = 1$.