Simultaneous Linear Equations

The procedure:

1. Make the coefficients of one variable EQUAL in magnitude.
2. ADD if signs are opposite, SUBTRACT if equal.
3. Solve the resulting single-variable equation.
4. Substitute back to find the other variable.

Example. $3x + 2y = 12$, $5x - 2y = 4$.

Coefficients of $y$: $+2$ and $-2$ → opposite. Add:

• $8x = 16$ → $x = 2$.
• Substitute: $3(2) + 2y = 12 \to y = 3$.

With multiplication. $2x + 3y = 13$, $4x + y = 11$.

Multiply 2nd by $3$ to align $y$: $12x + 3y = 33$. Subtract 1st from this: $10x = 20$ → $x = 2$. Substitute back: $4(2) + y = 11 \to y = 3$.

Sign rules:

• SAME sign coefficients → SUBTRACT.
• OPPOSITE sign coefficients → ADD.

Edexcel tip. Always state your strategy: 'multiplying equation 2 by 3 to align $y$-coefficients'. Mark schemes credit this clarity.

Useful when: one variable is already isolated (e.g. $y = ...$).

Method:

1. Solve one equation for one variable.
2. Substitute into the other equation.
3. Solve.
4. Substitute back to find the other variable.

Example. $y = 2x + 1$, $3x + y = 11$.

Sub $y = 2x + 1$ into 2nd: $3x + (2x + 1) = 11 \to 5x + 1 = 11 \to x = 2$.

Sub back: $y = 2(2) + 1 = 5$.

When to use substitution vs elimination.

Situation	Method
One variable already isolated	Substitution
Coefficients align cleanly	Elimination
Linear-quadratic	Substitution (must)

Worked qualitative. Why does substitution work for linear-quadratic?

• The linear gives $y$ in terms of $x$ (or vice versa).
• Substituting into the quadratic gives a single-variable quadratic.
• Solve for $x$, get the corresponding $y$ values.

Edexcel tip. Show the substitution explicitly. 'Substituting $y = 2x + 1$ into ...' earns the M1.

Form: one linear equation + one quadratic equation.

Linear: $y = mx + c$ (or rearrange to this form). Quadratic: $x^{2} + y^{2} = r^{2}$ (circle), $y = x^{2} + ...$, etc.

Method:

1. From the linear equation, express $y$ in terms of $x$.
2. Substitute into the quadratic.
3. Solve the resulting quadratic in $x$.
4. For each $x$, find the corresponding $y$.

Example. $y = x + 2$, $x^{2} + y^{2} = 10$.

Sub: $x^{2} + (x + 2)^{2} = 10$. Expand: $2x^{2} + 4x + 4 = 10 \to 2x^{2} + 4x - 6 = 0$. Divide by 2: $x^{2} + 2x - 3 = 0$. Factor: $(x + 3)(x - 1) = 0$. $x = -3$ or $x = 1$.

For each, find $y$:

• $x = -3 \to y = -3 + 2 = -1$. Pair: $(-3, -1)$.
• $x = 1 \to y = 1 + 2 = 3$. Pair: $(1, 3)$.

TWO solution pairs. Always state both as pairs.

Geometric interpretation. A line through a circle: 0, 1, or 2 intersection points. Two solutions = two intersections.

Worked qualitative. Why two solution pairs?

• A linear equation has one degree of freedom.
• A quadratic in two variables has many.
• Their intersection is finite: at most degree-of-quadratic many points.
• For a quadratic and a line: typically 2 intersections.

Edexcel tip. State BOTH pairs as $(x, y)$. Mark schemes deduct for stating only one.

Process:

1. Identify the unknowns. Define them with letters.
2. Set up TWO equations from the problem.
3. Solve simultaneously.
4. Check both unknowns make sense.

Example. Three pens + two books cost £$13$. Five pens + four books cost £$23$.

• Let pen = £$p$, book = £$b$.
• $3p + 2b = 13$.
• $5p + 4b = 23$.
• Multiply first by 2: $6p + 4b = 26$.
• Subtract second from this: $p = 3$.
• Substitute: $b = 2$.

So pen = £$3$, book = £$2$.

Common contexts:

• Cost / price problems (different items).
• Mixture / mixing problems.
• Speed / distance / time (current vs still water).
• Ages.

Worked qualitative. Two trains leave at the same time. Train A travels at $60$ km/h, train B at $80$ km/h. After how long are they $560$ km apart (going in opposite directions)?

• Let time = $t$ hours.
• Distance A: $60t$. Distance B: $80t$.
• Total apart: $60t + 80t = 140t = 560$.
• $t = 4$ hours.

(Single equation, but linear simultaneous in distance + time pattern.)

Edexcel tip. Always define variables explicitly. 'Let $p$ = price of a pen' earns the first M1.