NapkinCalc

Algebra 1B — Systems, Exponents & Factoring

Systems of equations

ELI5: one equation with two unknowns has infinitely many answers — but two equations pin down a single (x, y). Elimination is the cleanest by-hand method: line the equations up so that adding them makes one variable vanish.

Solve 2x + y = 8 and x − y = 1: add them and the y's cancel → 3x = 9 → x = 3, then back-substitute for y = 2.

xs=8+13x_{s} = \frac{8 + 1}{3} = 33 add the equations: (2x + y) + (x − y) = 9
ys=xs1y_{s} = x_{s} - 1 = 22 back-substitute into x − y = 1
✓ pass abs(2xs+ys8)<109\mathrm{abs}\left(2 \cdot x_{s} + y_{s} - 8\right) < 10^{-9} first equation holds
✓ pass abs(xsys1)<109\mathrm{abs}\left(x_{s} - y_{s} - 1\right) < 10^{-9} second equation holds

Elimination by hand is the understanding; solveSystem is the shortcut — give it the equations, the variable you want, and the list of unknowns.

xsys:=solveSystem(["2x+y=8","xy=1"],"x",["x","y"])x_{sys} := solveSystem(["2*x + y = 8", "x - y = 1"], "x", ["x", "y"]) = 33 the engine solves the pair for x
✓ pass xsys==xsx_{sys} == x_{s} matches the elimination by hand

Real-world hook: systems are everywhere two quantities trade off — mixing two coffee blends to a target price, splitting a trip between two speeds, balancing supply and demand.

Try it yourself: solve x + y = 7 and 2x − y = 8 for x. (Add the equations: the y's cancel.)

xyou=5x_{you} = 5 ✏️ Your turn: solve the system for x. The check substitutes y = 2x − 8 (from the 2nd equation) into x + y = 7 — so you must find the real x.
✓ pass abs(xyou+2xyou87)<109\mathrm{abs}\left(x_{you} + 2 \cdot x_{you} - 8 - 7\right) < 10^{-9} green when your x satisfies both equations