R5 — Equations

Goal: solve equations, inequalities and systems — and understand the one thing solve does that will surprise you.

One equation, one unknown

> solve(2x + 6 = 0, x)
x := -3.0

run this session

Why does that look like an assignment?

Because it is one. Read the echo again — x := -3.0, the exact shape R3 taught. solve does not merely report the root, it stores it, so the next line that mentions x uses it:

> x + 1
-2.0

run this session

That is convenient when you want to carry a result forward into further work, and a trap when you do not: a second solve would substitute the first answer before doing anything at all. This is the stale-binding problem R3 warned about, arriving from a command you did not think of as an assignment.

Clear it between problems:

> unset x
x unset
> solve(x^2 - 4 = 0, x)
x_1 := -2.0
x_2 := 2.0

run this session

Two roots, and notice they bound as x_1 and x_2 rather than x — so x itself is untouched. Proof, rather than assertion:

> unset x
x is not set

run this session

Nothing to clear. The habit to build is simple: when a solve answer looks wrong, check env first.

Symbolic coefficients

> solve(a*x + b = 0, x)
x := ((-1.0 * b) / a)
> unset x
x unset

run this session

That is what makes this a computer algebra system rather than a numerical solver: it rearranged the equation without ever knowing what a and b are.

When there is no closed form

> solve(cos(z) = z, z)
z := 0.73909
> unset z
z unset

run this session

No formula exists for that root, so the answer is numeric — and which kind you got is visible in the answer itself rather than hidden behind a flag.

Naming an equation

An equation is an ordinary value, so you can store one and solve it later:

> eq := x^2 - 5x + 6 = 0
eq := (((x ^ 2.0) + (-1.0 * (5.0 * x))) + 6.0) = 0.0
> solve(eq, x)
x_1 := 2.0
x_2 := 3.0
> unset x_1
x_1 unset
> unset x_2
x_2 unset

run this session

Inequalities

The answer to an inequality is a region, and the language can say so:

> solve(2x + 6 > 0, x)
(x > -3.0)
> solve(x^2 - 4 > 0, x)
((x < -2.0) or (x > 2.0))

run this session

The second is a union, written with or — an ordinary connective here, not a special kind of object. A bounded region comes back as an and, and you can pose a combination directly:

> solve(x^2 - 4 < 0, x)
((x > -2.0) and (x < 2.0))
> solve(2x > 2 and x < 5, x)
((x > 1.0) and (x < 5.0))

run this session

Note there is no := in any of those: a region is not a value to assign, so nothing was stored. Only a definite answer binds.

Systems

solveSystem takes a row of equations, then the unknowns in order:

> solveSystem([[x + y = 3, x - y = 1]], x, y)
[[x = 2.0, y = 1.0]]

run this session

The [[...]] brackets are the matrix literal — one row — which is also the shape the answer comes back in. R6 is about those.

Equations are values too

An equation evaluates as soon as everything in it is known:

> 2 + 2 = 4
true
> sin(pi) = 0
true

run this session

The second is true because equality here is tolerance-based at the display precision. sin(pi) is not exactly zero in floating point — π is not exactly π — and answering false would be technically correct and practically useless. R8 is about getting the exact answer when you want it.

Try it yourself

A rectangle has perimeter 20 and area 24. Find its sides.

Answers

The obvious attempt is a system: ```text > solveSystem([[2w + 2h = 20, w * h = 24]], w, h) solveSystem([[((2.0 * w) + (2.0 * h)) = 20.0, (w * h) = 24.0]], w, h) ``` *[run this session](https://lycogrypho.github.io/Leonardo/app/#s=precision%205%0Acolors%20dark%0Apretty%20off%0Alatex%20off%0Anames%20off%0Alogic%20minmax%0Alogic%20symmetric%20off%0Aexact%20precision%2030%0Aexact%20off%0Aeq%20%3A%3D%20(((x%20%5E%202.0)%20%2B%20((-1.0)%20*%20(5.0%20*%20x)))%20%2B%206.0)%20%3D%200.0)* Declined — the unchanged-answer shape from [R4](/Leonardo/lessons/r4.html). **`solveSystem` is linear only**, and `w * h` is not, so it did not guess. Substitute by hand (`h = 10 - w`) and solve the quadratic that remains: ```text > solve(w * (10 - w) = 24, w) w_1 := 4.0 w_2 := 6.0 ``` *[run this session](https://lycogrypho.github.io/Leonardo/app/#s=precision%205%0Acolors%20dark%0Apretty%20off%0Alatex%20off%0Anames%20off%0Alogic%20minmax%0Alogic%20symmetric%20off%0Aexact%20precision%2030%0Aexact%20off%0Aw_1%20%3A%3D%204.0%0Aw_2%20%3A%3D%206.0%0Aeq%20%3A%3D%20(((x%20%5E%202.0)%20%2B%20((-1.0)%20*%20(5.0%20*%20x)))%20%2B%206.0)%20%3D%200.0)* Sides 4 and 6 — the two roots are the same rectangle, seen twice.

What you learned

solve binds its answer — one root as the variable, several as x_1, x_2, … Use unset between problems, or the next solve starts from the last one.
It handles linear and quadratic exactly, symbolic coefficients included, and falls back to numeric roots when no formula exists.
An inequality’s answer is a region, written with and / or in the same language you asked in — and regions do not bind.
solveSystem([[...]], x, y) is linear only, and declines rather than guessing.
Equality between numbers is tolerant, so sin(pi) = 0 is true.

Previous: R4 — Calculus · Next: R6 — Matrices · All lessons