Simultaneous Equations — Substitution
Grade 6When substitution works best
The substitution method shines when one equation gives you a letter already isolated — e.g. y = 2x + 1. You simply plug that expression into the other equation in place of y.
It is also the only method that works cleanly when one equation is quadratic (like x² + y² = 25 paired with y = x + 1).
Worked example — linear pair by substitution
EXAMPLE
Solve: y = 3x − 2 and 4x + 2y = 20.
Substitute y = 3x − 2 into the second equation: 4x + 2(3x − 2) = 20. Expand: 4x + 6x − 4 = 20. Simplify: 10x = 24, x = 2.4. Find y: y = 3(2.4) − 2 = 7.2 − 2 = 5.2. Answer: x = 2.4, y = 5.2.
Worked example — linear and quadratic pair
EXAMPLE
Solve: y = x + 1 and x² + y² = 25.
Substitute y = x + 1: x² + (x + 1)² = 25. Expand: x² + x² + 2x + 1 = 25. Simplify: 2x² + 2x − 24 = 0, so x² + x − 12 = 0. Factorise: (x + 4)(x − 3) = 0. So x = −4 or x = 3. If x = −4: y = −3. If x = 3: y = 4. Solutions: (3, 4) and (−4, −3).
Two solutions are normal for non-linear pairs
A line can intersect a circle or parabola at two points, giving two pairs of answers. Present both clearly — one answer on its own is only half marks.
Try this one
Solve the simultaneous equations
\[ 4x + 3y = 27 \]
\[ 4x + y = 17 \]
Nova's hint:
Try to Solve it with Nova?
This is solving simultaneous equations by elimination.
Ask yourself: 'the \( 4x \) terms are the same, so what does subtracting leave?' For example, subtracting equal \( 4x \) terms removes \( x \) entirely.
