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ALGEBRA · HLIntroduction
Algebra · Ordinary Level

Simultaneous Equations

Ordinary Level  ·  Class 7 of 8  ·  Tap NEXT to begin

Section 1 of 2

Elimination — Add or Subtract

One equation in two unknowns has infinitely many solutions. To pin down a single pair $(x, y)$ you need a second equation. The two lines cross at that solution.
The recipe
2 unknowns, 2 equations: eliminate one letter, solve for the other, back-substitute. If the signs on the matching letter are different, add; if the signs are the same, subtract.

Worked example — different signs — add

Solve $x + y = 3$ and $x - y = 1$.
Add the equations: $2x = 4 \Rightarrow x = 2$
Sub into the first: $2 + y = 3 \Rightarrow y = 1$
$(x, y) = (2, 1)$

Worked example — another, add

Solve $2x + y = 3$ and $3x - y = 2$.
Add: $5x = 5 \Rightarrow x = 1$
$2(1) + y = 3 \Rightarrow y = 1$
$(x, y) = (1, 1)$
You try
Solve $5x + y = 16$ and $3x + y = 10$.
The $y$ signs are the same — subtract the equations to cancel $y$.
Subtract: $2x = 6 \Rightarrow x = 3$
$5(3) + y = 16 \Rightarrow y = 1$
$(x, y) = (3, 1)$
Section 2 of 2

The Multiplier Method

Most questions don’t hand you matching coefficients — you make them match. Pick the letter to eliminate and multiply each equation so that letter has the same number, different sign.
Make them match
Multiply each equation by whatever it takes so one letter’s coefficients match. Then add (or subtract) to eliminate it.

Worked example — multiply both, then add

Solve $5x + 3y = 19$ and $4x - 5y = -7$.
$\times 5$: $\ 25x + 15y = 95$;  $\times 3$: $\ 12x - 15y = -21$
Add: $37x = 74 \Rightarrow x = 2$
$5(2) + 3y = 19 \Rightarrow 3y = 9 \Rightarrow y = 3$
$(x, y) = (2, 3)$
You try
Solve $3x + 2y = 13$ and $4x - 3y = 6$.
Multiply the first by $3$ and the second by $2$: the $y$ coefficients become $+6$ and $-6$. Then add.
$\times 3$: $\ 9x + 6y = 39$;  $\times 2$: $\ 8x - 6y = 12$
Add: $17x = 51 \Rightarrow x = 3$
$3(3) + 2y = 13 \Rightarrow 2y = 4 \Rightarrow y = 2$
$(x, y) = (3, 2)$

That’s Class 7.

Eliminate a letter by adding or subtracting — and when nothing matches, multiply to make it match. Class 8: quadratic equations.

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