Simultaneous equations · Linear & simultaneous equations

One equation with two unknowns, like $x + y = 10$ , has endless solutions: $1$ and $9$ , $4$ and $6$ , $2.5$ and $7.5$ … A second equation narrows it down to the one pair that works for both at the same time. “Simultaneous” means “at the same time”, which is where the name comes from.

What the answer means

The graph of a linear equation in $x$ and $y$ is a straight line, and every point on the line is a solution. Two lines usually cross at exactly one point. That point is the only pair $(x, y)$ on both lines.

Two lines, one answerMove the lines

y = x + 1 and y = −2x + 4

(1, 2)where they crossx = 1, y = 2the solution

gradient of line 1 = 1line 1 cuts the y-axis at = 1gradient of line 2 = −2line 2 cuts the y-axis at = 4

Every point on a line satisfies its equation. Only the crossing point, (1, 2), is on both lines, so it's the only pair of values that makes both equations true. Check: 1 × 1 + 1 = 2.

Make the two gradients (slopes) equal and watch what happens to the solution.

The solution is the crossing pointThe pair (x, y) that lies on both lines

Drawing is slow and only gives rough answers, so we usually solve with algebra.

Method 1: elimination

Make the number in front of one letter the same in both equations. Then add or take away the equations to eliminate (remove) that letter.

\begin{aligned} 3x - 2y &= 10 \quad (1)\\ x + 3y &= 7 \quad (2) \end{aligned}

Multiply (2) by 3, so both have $3x$ : ${3x + 9y = 21}$ . Call this (3).
Take (1) from (3): ${11y = 11}$ , so ${y = 1}$ .
Put ${y = 1}$ into (2): ${x + 3 = 7}$ , so ${x = 4}$ .

Elimination, step by stepPick a pair, then press Next

Line them up
3x + 2y = 17
5x − 2y = 7

3x + 2y = 17 and 5x − 2y = 7solve

Line the equations up, x under x and y under y. The y terms are +2y and −2y: the same size with opposite signs. Press Next.

Method 2: substitution

Rearrange one equation so one letter is on its own. That letter is now the subject. Then put what it equals into the other equation, in brackets. This works best when a letter is already almost alone.

Substitution, step by stepPick a pair, then press Next

Start
y = 2x − 1
3x + 2y = 12

y = 2x − 1 and 3x + 2y = 12solve

The first equation already gives y on its own, so put it in place of y in the second. Press Next.

Worked example · WAEC 2020

WAEC 2020 · Paper 1 · Q10

Solve $3x - 2y = 10$ and $x + 3y = 7$ simultaneously.

Pick the easier letter
In $x + 3y = 7$ , $x$ has no number in front of it.
Think first. Which letter can you get on its own in one move?
Make it the subject
$x = 7 - 3y$
Substitute into the other equation
$\begin{aligned} 3(7 - 3y) - 2y &= 10 \\ 21 - 9y - 2y &= 10 \\ -11y &= -11 \end{aligned}$
so $y = 1$ .
Think first. Expand and solve for $y$ .
Find the other letter, then check
$x = 7 - 3(1) = 4$ . Check in the first equation: $3(4) - 2(1) = 10$ ✓. The answer is D.

Equations with fractions like 1/x

If $\frac1x$ and $\frac1y$ appear, treat them as the unknowns. Let $a = \frac1x$ and $b = \frac1y$ , solve for $a$ and $b$ , then turn them back: $x = \frac1a$ .

One linear, one quadratic

When one equation is a quadratic, make one letter the subject of the linear equation and substitute it into the quadratic. This is taught in quadratics.

More: one linear, one quadratic

More: simultaneous equations in disguise (powers, logs, matrices)

Your turn

WAEC 2023 · Paper 1 · Q39

Linear & simultaneous equations

If $2x - 3y = -11$ and $3x + 2y = 3$ , evaluate $(y - x)^2$ .

Worked solution (try it first)

Make the $y$ terms opposite: multiply the first by 2 and the second by 3, giving $4x - 6y = -22$ and $9x + 6y = 9$ .
Add them: $13x = -13$ , so $x = -1$ .
Put $x = -1$ into $3x + 2y = 3$ : $-3 + 2y = 3$ , so $y = 3$ .
So $y - x = 3 - (-1) = 4$ , and $(y - x)^2 = 16$ , option C.

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