Visualgebra

Drawing a relationship

A rule connecting x and y is not a number — it is a whole line of points at once.

Every lesson so far has plotted one point at a time. This one plots a rule.

A fixed line of points, all satisfying y=x+1y = x + 1, and one draggable point P. A live reading checks whether P’s own coordinates satisfy the same rule.

A labelled x and y axis with a line of fixed points all satisfying y = x + 1, and one draggable point P. A live reading states what y would need to be for P to satisfy the rule too, and whether it does. This needs JavaScript switched on.

Drag P anywhere. It only lands exactly on the fixed line where the rule is true.

Try to find a spot for P that’s close to the line but doesn’t count — the reading will show exactly how far off the rule it is, even by a fraction.

A rule connecting two varying quantities — here, xx and yy — is a relationship. xx is not one fixed unknown to solve for here, the way it is in an equation like 2x+1=72x + 1 = 7; it is a number that varies, with yy varying alongside it according to the rule.

One rule, tested against every point at once

y=x+1y = x + 1 is not a statement about one pair of numbers. It is a test any pair can be run through: put in a value for xx, and the rule says exactly what yy has to be for that pair to belong to the relationship. Every point on the fixed line passed that test; P only joins them when it does too.

(x,y) belongs to y=x+1    yx=1(x, y) \text{ belongs to } y = x + 1 \iff y - x = 1

This is the bridge the rest of algebra crosses on: an equation with two letters in it is not something to “solve” for a single answer, it describes a whole collection of pairs. The next unit gives that collection a name (a function), a picture (a graph), and a shape (a straight line, from y=mx+cy = mx + c), but the idea itself is already complete here.

Testing points against the rule

Does the point (4,5)(4, 5) satisfy y=x+1y = x + 1?

x+1=4+1=5x + 1 = 4 + 1 = 5, and that is exactly the point’s yy. Yes, it satisfies the relationship.

Does the point (3,3)(3, 3) satisfy y=x+1y = x + 1?

x+1=3+1=4x + 1 = 3 + 1 = 4, but the point’s yy is 3, not 4. No — it is 1 short.

Find a point, other than (2,3)(2, 3), that satisfies y=x+2y = x + 2.

Any pair where yy is exactly x+2x + 2 works, for example (0,2)(0, 2): x+2=0+2=2x + 2 = 0 + 2 = 2, which matches.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.