A19, A21 - Simultaneous linear equations

Eliminating an unknown

Two notebooks and three pens cost £9. Three notebooks and two pens cost £11. Can we recover both prices? One total allows many price pairs, so we need to use both purchases together.

For example, $x+y=7$ allows $(0,7),(2,5)$ and $(7,0)$. A second independent relationship may select just one pair. Elimination adds or subtracts complete equations so that one unknown cancels.

Change only the sign of the $y$-term in the second equation. This changes the useful operation from subtraction to addition.

If coefficients do not match, multiply every term of one or both equations. For $2x+3y=13$ and $3x-2y=0$, multiply the first by 2 and the second by 3:

Checking gives $4+9=13$ and $6-6=0$. Matching the $x$-coefficients instead would also work.

A pair must satisfy both originals. Finding one value is only the first half of the solution.

Substitution between equations

Substitution replaces an unknown by an equal expression from the other equation. It is especially useful when one unknown is already isolated.

The replacement uses both conditions without guessing a value. After finding one unknown, return to an original equation to find the other.

Change only the total in the second equation, from 16 to 21. The substitution expression stays unchanged.

Sometimes rearrange first. For $x+y=5$ and $2x-3y=0$, write $x=5-y$:

Both $3+2=5$ and $2(3)-3(2)=0$ check. Substitution suits an isolated unknown; elimination suits easily matched coefficients. Either can produce exact fractions: $x+y=3$, $x-y=2$ give $x=\frac52,\ y=\frac12$.

If two numbers are integers, their sum and difference have the same parity: subtracting the difference from the sum gives $2y$, an even number. That explains why an odd sum of 3 and an even difference of 2 could not produce two integers here.

If all unknown terms cancel, inspect what remains. The equations $y=2x+1$ and $2y=4x+2$ give a true identity, so every point on that line works. The equations $y=2x+1$ and $y=2x+4$ give $1=4$, so no pair works.

Keep the full replacement expression together until you have expanded correctly.

Intersections and numbers of solutions

Each linear equation describes a straight line. A common point lies on both, so its two coordinates form a simultaneous solution.

For $x+y=5$, use $y=5-x$, with points $(0,5),(2,3),(5,0)$. For $y=2x-1$, use $(0,-1),(2,3),(3,5)$. Plot both on the same axes with the same scales and draw each line with a ruler.

[DIAGRAM: asset_name: Common point of two linear equations; asset_slug: 1ma1_m29_linear_intersection; caption: The common point (2, 3) satisfies both straight-line equations.; description: A blue line y = 5 − x and clay line y = 2x − 1 meet at a sage point labelled (2, 3). Dashed guides project to x = 2 and y = 3 on labelled coordinate axes.]