Elimination, substitution and the graphical method all find the same solution. Elimination and substitution are exact; the graphical method also shows the two lines but usually gives an estimate. Choose the method that makes the working shortest. Why both algebraic methods work: each turns two equations in two unknowns into one equation in one unknown. Elimination does it by adding or subtracting equations; substitution does it by replacing one unknown with an equal expression. **What the equations look like** | **A good choice** | **Reason** one unknown is already the subject, as in y=7x-2 | substitution | the expression is ready to substitute an unknown has coefficient 1 or -1 | substitution or elimination | it is easy to make it the subject one unknown has the same or opposite coefficients | elimination | add or subtract straight away all coefficients are larger numbers | elimination after multiplying | substitution would bring in fractions a graph is asked for, or an estimate is enough | graphical method | the intersection gives the solution Method - Look at the coefficients and decide which method fits, using the table. - Solve, showing each step. - Check the pair in both original equations, not in a rearranged or multiplied version. Example Solve 4x+y=5 (1) and 7x-3y=23 (2). No unknown is the subject, but y has coefficient 1 in (1), so both methods are short. Compare them. Substitution: from (1), y=5-4x. 7x-3(5-4x)=23 19x-15=23 x=2 Elimination: (1) x 3 gives 12x+3y=15. (12x+3y)+(7x-3y)=15+23 19x=38 x=2 Both give x=2; then y=5-4(2)=-3. Check: in (1), 4(2)+(-3)=5; in (2), 7(2)-3(-3)=14+9=23. Why check both: the pair (5,4) satisfies (2), because 7(5)-3(4)=23, but not (1), because 4(5)+4=24 5. Checking only one equation would let a wrong answer pass.

Choosing a method and checking

Secondary 2

Elimination, substitution and the graphical method all find the same solution. Elimination and substitution are exact; the graphical method also shows the two lines but usually gives an estimate. Choose the method that makes the working shortest.

Why both algebraic methods work: each turns two equations in two unknowns into one equation in one unknown. Elimination does it by adding or subtracting equations; substitution does it by replacing one unknown with an equal expression.

What the equations look like	A good choice	Reason
one unknown is already the subject, as in $y=7x-2$	substitution	the expression is ready to substitute
an unknown has coefficient $1$ or $-1$	substitution or elimination	it is easy to make it the subject
one unknown has the same or opposite coefficients	elimination	add or subtract straight away
all coefficients are larger numbers	elimination after multiplying	substitution would bring in fractions
a graph is asked for, or an estimate is enough	graphical method	the intersection gives the solution

Method

Look at the coefficients and decide which method fits, using the table.
Solve, showing each step.
Check the pair in both original equations, not in a rearranged or multiplied version.

Example

Solve $4x+y=5$ $(1)$ and $7x-3y=23$ $(2)$. No unknown is the subject, but $y$ has coefficient $1$ in $(1)$, so both methods are short. Compare them.

Substitution: from $(1)$, $y=5-4x$.

$\begin{aligned} 7x-3(5-4x)&=23 \\ 19x-15&=23 \\ x&=2 \end{aligned}$

Elimination: $(1)\times3$ gives $12x+3y=15$.

$\begin{aligned} (12x+3y)+(7x-3y)&=15+23 \\ 19x&=38 \\ x&=2 \end{aligned}$

Both give $x=2$; then $y=5-4(2)=-3$. Check: in $(1)$, $4(2)+(-3)=5$; in $(2)$, $7(2)-3(-3)=14+9=23$.

Why check both: the pair $(5,4)$ satisfies $(2)$, because $7(5)-3(4)=23$, but not $(1)$, because $4(5)+4=24\ne5$. Checking only one equation would let a wrong answer pass.

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