Simultaneous Equations

One unknown with one equation — solving linear equations covered that. Real life, though, often hands you two unknowns at once: "How much is an apple? And an orange?" One clue alone leaves infinitely many answers; two clues together pin the answer down.

Two unknowns, two clues

Consider this pair:

$x + y = 10, \qquad x - y = 2$

What are $x$ and $y$ ? The first equation alone has infinitely many solutions (1 and 9, 5 and 5, 6.5 and 3.5...); so does the second. One equation cannot settle it — only together do they lock in a single pair. That is what a "system" means, and what "simultaneous" means: the pair you find must satisfy both equations at once.

Substitution: swap an unknown for an expression

Substitution, step by step

Solve $x + y = 10$ , $x - y = 2$ .

From the second equation, write $x$ in terms of $y$ : $x = y + 2$ ;
Substitute into the first: $(y + 2) + y = 10$ , that is $2y + 2 = 10$ ;
Subtract 2 from both sides, then divide by 2: $y = 4$ ;
Substitute back: $x = y + 2 = 6$ .

The solution is $x = 6$ , $y = 4$ . Two unknowns, but we eliminated one first and landed on a familiar one-unknown equation.

Elimination: unknowns vanish by adding

A faster road is simply to add the two equations: on the left, $x + x$ makes $2x$ while $y$ and $-y$ cancel; on the right, $10 + 2 = 12$ :

$2x = 12 \quad\Rightarrow\quad x = 6$

Substitute $x = 6$ into either equation to get $y = 4$ . You could subtract instead: $(x + y) - (x - y) = 10 - 2$ gives $2y = 8$ , so $y = 4$ . Same system, same answer — substitution and elimination are two roads to one destination. Pick whichever suits the coefficients.

In practice: apples and oranges

A market equation system

3 apples plus 2 oranges cost 13; 1 apple plus 1 orange costs 5. What does each cost?

Let an apple be $a$ and an orange $o$ : $3a + 2o = 13$ and $a + o = 5$ .

From the second, $o = 5 - a$ ; substitute: $3a + 2(5 - a) = 13$ ;
Expand: $3a + 10 - 2a = 13$ , so $a + 10 = 13$ and $a = 3$ ;
Then $o = 5 - 3 = 2$ . Apples cost 3, oranges cost 2.

A mental shortcut: double "1 apple 1 orange" into "2 apples 2 oranges for 10" and compare with "3 apples 2 oranges for 13" — the extra is exactly one apple, worth 3. That is elimination done in your head.

Check against both

After solving, substitute the pair back into both equations: $6 + 4 = 10$ holds, $6 - 4 = 2$ holds — only then are you done. One equation is not enough; plenty of pairs satisfy just one.

Check yourself

Quick quiz

1. x+y=10 and x−y=2. What is x?
2. Why does solving for two unknowns take two equations?
3. 3 apples and 2 oranges cost 13; 1 apple and 1 orange cost 5. Price of one apple?