Two equations in two unknowns.

Step 1 of 4

A point on both lines

An equation with two unknowns, such as $y = x + 1$⁠, has many solutions: every pair $(x, y)$⁠ that satisfies it. Mark them all on the board and they make a straight line. So each of the two equations

$y = x + 1 \qquad y = -x + 5$$y =$$x + 1 \qquad y =$$-x + 5$⁠

is a line, and a pair that satisfies both sits on both lines at once: where they meet. That is the solution of the system of equations. A point is written $(x, y)$⁠: first the distance across, then the height.

Step 2 of 4

Substitution

When one equation already says what $y$⁠ is, substitute that expression for $y$⁠ in the other equation. In the system

$y = 2x - 1 \qquad x + y = 8$$y =$$2x - 1 \qquad x + y =$$8$⁠

put $2x - 1$⁠ in place of $y$⁠: $x + (2x - 1) = 8$⁠. That leaves an equation with one unknown, which you already know how to solve. On the board the second equation is written $y = -x + 8$⁠, the same equation with $y$⁠ isolated.

Step 3 of 4

Adding the equations

Sometimes it is easier to add the equations: left side to left side, right side to right side. If both are true, so is their sum. In the system

$x + y = 5 \qquad x - y = 1$$x + y =$$5 \qquad x - y =$$1$⁠

the $y$⁠ appears once with a plus and once with a minus, so adding makes it disappear. On the board they are written $y = -x + 5$⁠ and $y = x - 1$⁠.

Step 4 of 4

Parallel lines

Not every system has one solution. In the two lines on the board, $y = 2x + 1$⁠ and $y = 2x - 3$⁠, $y$⁠ grows at the same rate: $2$⁠ for every step to the right.