Q.L-algebra · Algebra: domains, systems, inequalities and functions

Write restrictions before manipulating an equation. A denominator must be nonzero; a real even root needs a nonnegative argument. Cancelling factors preserves the expression only on its original domain. Squaring can introduce candidates because a square loses sign information; verify every candidate in the unsquared relation.

Solving a system requires the same ordered pair to satisfy all relations. Substitution or elimination can expose one, no or infinitely many linear solutions. For an inequality, multiplication or division by a negative reverses direction. For a factored quadratic inequality, order roots and test interval signs, including endpoints only when equality is allowed.

A function assigns one output to each permitted input. Evaluate the inner function first in a composition, preserving domain restrictions. Read a slope as change in output per input unit and an intercept as the output at input zero only when that input is meaningful. A verbal model should name variables and keep quantities in compatible units.

For Quantitative Comparison, preserve all allowed values. A variable with x²=4 could be −2 or 2 unless constrained. One counterexample can reject a universal relationship; several favourable substitutions cannot prove it for every real value. Algebraic structure can establish a relationship without exhaustive numerical sampling.

For √(x+2)=x, x≥0. Squaring gives x²−x−2=0, candidates 2 and −1. Only 2 satisfies the original. For (x−1)(x−4)≤0, an upward quadratic is nonpositive on 1≤x≤4, including roots. In the system x+y=10, x−y=2, adding gives 2x=12; x=6 and y=4. For f(t)=2t+1 and g(x)=x², f(g(3))=2·9+1=19, applying g first.

A transformed equation supplies candidates, not automatic original solutions. Negative inequality multipliers and strict endpoints need explicit checks.

Solve $(x^2-4)/(x-2)=5$ on its original real domain. Solve $(x+1)(x-3)\le0$ and find $f(g(2))$ for $f(t)=t^2$, $g(x)=x-3$.

Model and reasoning

The rational equation excludes x=2. For allowed x it reduces to x+2=5, giving x=3, which substitutes correctly. The quadratic is nonpositive between its roots, including them: $-1\le x\le3$. The composition applies g first: $g(2)=-1$ and $f(g(2))=1$. Cancelling a factor does not restore excluded inputs.

For real a, classify the system $x+y=4$, $2x+ay=8$ as having one, none or infinitely many solutions. Compare with the second equation $2x+ay=9$. Give solutions in every consistent case.

Check after attempting

Subtract twice the first equation. In the first system $(a-2)y=0$: for a not equal to two, y=0,x=4, uniquely. For a=2, both equations are the same relation, giving every pair $(4-t,t)$. There is no inconsistent case in this first system. With right side nine, $(a-2)y=1$. For a not equal to two, $y=1/(a-2)$ and $x=4-1/(a-2)$, uniquely. For a=2 it demands 0=1, so no solution. A zero coefficient row means inconsistency only when the right side is nonzero.