6.03c - Simultaneous linear equations

What a simultaneous solution means

Two notebooks and a pen cost £7. One notebook and a pen cost £4. Can you find both prices without knowing either price first?

Let $n$ be a notebook's price and $p$ a pen's price, both in pounds. The two receipts give $2n+p=7$ and $n+p=4$. They describe the same prices: changing $n$ between the equations would describe a different shop.

Consider $x+y=7$ and $x-y=1$. The pair $x=5$, $y=2$ satisfies the first equation, because $5+2=7$, but fails the second, because $5-2=3$. The pair $x=4$, $y=3$ satisfies both: $4+3=7$ and $4-3=1$. Only the second pair is a simultaneous solution.

Two linear equations do not always determine one pair. For $x+y=7$ and $2x+2y=14$, the second equation repeats the first, so every pair satisfying the first also satisfies the second. But $x+y=7$ and $2x+2y=15$ cannot both be true: doubling the first would give fourteen, not fifteen. These have infinitely many solutions and no solutions respectively.

Test both equations rather than choosing the first pair that fits one.

Eliminating a variable

Elimination combines two true equations so that one variable cancels. Subtract equations with equal coefficients; add equations with opposite coefficients.

This works because equal amounts remain equal when added or subtracted. If $A=B$ and $C=D$, then $A-C=B-D$. Apply subtraction to every term of the second equation, including any negative terms.

Change only the sign of the $y$-term in the second equation. Adding now cancels the variable.

When coefficients do not yet match, multiply a whole equation first. For $2x+3y=13$ and $x+y=5$, doubling the second gives $2x+2y=10$. Subtracting this from the first gives $y=3$, then $x=2$. Both sides and every term must be multiplied. Sometimes you must scale both equations to reach a common multiple.

This additional case changes the preparation, not the cancellation rule: use a common multiple to produce equal or opposite coefficients before combining.

If elimination leaves $0=0$, the two equations may be equivalent and have infinitely many solutions. If it leaves $0=5$, the equations contradict each other. Do not divide by zero to manufacture a value.

Scale both equations to make a matching pair of coefficients.