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Linear & simultaneous equations · Lesson 2 of 3

Simultaneous equations

Two equations, two unknowns: what the answer means on a graph, and how elimination and substitution find it.

15 minYou should already know: Expressions, formulae & change of subject
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One equation with two unknowns, like x+y=10x + y = 10, has endless solutions: 11 and 99, 44 and 66, 2.52.5 and 7.57.5… A second equation narrows it down to the one pair that works for both at the same time. That’s why they’re called simultaneous equations.

What the answer means

Every linear equation in xx and yy is a straight line: its solutions are the points on the line. Two lines usually cross at exactly one point, and that point is the only pair (x,y)(x, y) on both lines.

Two lines, one answerMove the lines

y = x + 1 and y = −2x + 4

−6−5−4−3−2−1123456−8−6−4−22468xy(1, 2)
(1, 2)where they crossx = 1, y = 2the solution
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 equal and see what happens to the solution.

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

Drawing is slow and only gives rough answers, so we solve them 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 that letter.

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

Multiply (2) by 3 so both have 3x3x: 3x+9y=213x + 9y = 21 (3). Take (1) from (3): 11y=1111y = 11, so y=1y = 1. Put y=1y = 1 into (2): x+3=7x + 3 = 7, so x=4x = 4.

Method 2: substitution

Rearrange one equation to make one letter the subject, then put that into the other equation. It’s best when a letter already stands almost alone.

Worked example · WAEC 2020

WAEC 2020 · Paper 1 · Q10

Solve 3x−2y=103x - 2y = 10 and x+3y=7x + 3y = 7 simultaneously.

  1. Pick the easier letter

    In x+3y=7x + 3y = 7, xx has no number in front of it.

    Think first. Which letter can you get on its own in one move?

  2. Make it the subject

    x=7−3yx = 7 - 3y

  3. Substitute into the other equation

    3(7−3y)−2y=10⇒21−9y−2y=10⇒−11y=−113(7 - 3y) - 2y = 10 \quad\Rightarrow\quad 21 - 9y - 2y = 10 \quad\Rightarrow\quad -11y = -11

    so y=1y = 1.

    Think first. Expand and solve for yy.

  4. Find the other letter, then check

    x=7−3(1)=4x = 7 - 3(1) = 4. Check in the first equation: 3(4)−2(1)=103(4) - 2(1) = 10 ✓. The answer is D.

Equations with fractions like 1/x

If 1x\frac1x and 1y\frac1y appear, treat them as the unknowns. Let a=1xa = \frac1x and b=1yb = \frac1y, solve for aa and bb, then turn them back: x=1ax = \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

If 2x−3y=−112x - 3y = -11 and 3x+2y=33x + 2y = 3, evaluate (y−x)2(y - x)^2.

Worked solution (try it first)
  1. Make the yy terms opposite: multiply the first by 2 and the second by 3, giving 4x−6y=−224x - 6y = -22 and 9x+6y=99x + 6y = 9.
  2. Add them: 13x=−1313x = -13, so x=−1x = -1.
  3. Put x=−1x = -1 into 3x+2y=33x + 2y = 3: −3+2y=3-3 + 2y = 3, so y=3y = 3.
  4. So y−x=3−(−1)=4y - x = 3 - (-1) = 4, and (y−x)2=16(y - x)^2 = 16, option C.

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