EASA / CAR Part-66 · Module 1 · 1.2 Algebra · Cat B1/B2

Straight-Line Graphs & Simultaneous Equations

Objective: y = mx + c is a recipe for a line, where m tilts it and c lifts it. Set two lines and the point where they cross is the solution to both equations at once, which is what solving simultaneously means.

Two lines and their crossing point

Line 1 gradient m 1.0

Line 1 intercept c 1

Line 2 gradient m -0.5

Line 2 intercept c 4

Solution x

Solution y

m1 − m2

Status

y = mx + c · m = rise / run · the solution is where m₁x + c₁ = m₂x + c₂

m is the gradient and c is where the line cuts the y axis. Gradient is rise over run: how far the line climbs for every one step to the right. A negative m means the line falls, and m = 0 makes it flat.

Two equations, two unknowns, one crossing point. Every point on a line satisfies that line's equation, so the point sitting on both lines satisfies both equations. That is exactly what a simultaneous solution is, and it is why the graphical method works.

Parallel lines never meet, so there is no solution. Give both lines the same gradient and the crossing point disappears. Algebraically you end up with a statement like 3 = 5, which is the same fact stated differently.

Try this: Set both gradients to the same value. What happens to the solution, and what does that mean?
Try this: Make line 1 y = 2x + 1 and line 2 y = −x + 4. Solve it on paper, then check the readout.
Try this: Change only c on one line. Does the crossing point move along the other line?