Solving Equations
Pre-AlgebraSolving an equation means finding the value(s) of the variable that make the equation true, using inverse operations to isolate the variable.
Definition
Solving an equation means finding the value of the variable that makes both sides equal. You use opposite (inverse) operations to peel away what is attached to the variable until it stands alone.
Example
Solve $x - 3 = 8$. Add $3$ to both sides: $x = 11$. Check: $11 - 3 = 8$. Correct.
Key Insight
Whatever you do to one side of the equation you must do to the other side, just like keeping a balance scale even.
Definition
Solving an equation is the process of applying inverse operations systematically to isolate the variable. For linear equations, this involves undoing operations in reverse order of the order of operations (PEMDAS) while maintaining equality on both sides.
Example
Solve $3(x + 2) = 21$. Divide both sides by $3$: $x + 2 = 7$. Subtract $2$ from both sides: $x = 5$. Check: $3(7) = 21$. Correct.
Key Insight
The method of applying the same operation to both sides preserves equality. This is formally the substitution property of equality: if $a = b$, then $a + c = b + c$.
Definition
Solving an equation over a domain $D$ is finding the solution set $\{x \in D : f(x) = g(x)\}$. For linear equations over a field, Gaussian elimination solves systems efficiently in $O(n^3)$ time. Nonlinear equations may require numerical methods (Newton-Raphson iteration) when closed-form solutions do not exist.
Example
Newton-Raphson for solving $f(x) = 0$: $x_{n+1} = x_n - f(x_n)/f'(x_n)$. Starting at $x_0 = 2$ for $f(x) = x^2 - 2$: $x_1 = 2 - (4-2)/4 = 1.5$, $x_2 = 1.5 - (2.25-2)/3 = 1.4167$, converging to $\sqrt{2} = 1.4142\ldots$
Key Insight
Newton-Raphson converges quadratically near a simple root, meaning the number of correct decimal digits roughly doubles with each iteration. This makes it the gold standard for fast root-finding in numerical analysis.