Equations as balance: find the unknown without losing solutions

Three identical items plus delivery costing 40 hryvnias total 250 hryvnias. What does one item cost? Let $x$ be its price in hryvnias. The bill gives $3x+40=250$ , hence $x=70$ . To reach the answer, we change both sides alike and check whether each step can be reversed.

A variable names the unknown; domain restrictions determine where the notation makes sense. Here algebra permits all real $x$ , while a price requires a nonnegative value.

A solution makes the equality true

An equation is an equality containing unknowns. A solution of a one-variable equation is an allowed number whose substitution makes it true. For our bill, $3\cdot70+40=250$ ; 60 fails because it gives 220.

Equivalent equations have exactly the same solution set. Keeping that set requires more than obtaining a true new equality from the old one: we must also be able to go backwards.

Subtract 40 from both sides: $3x=210$ . Divide both by the nonzero number 3: $x=70$ . Both steps reverse by multiplying by 3 and adding 40, so no solution was lost or added.

Add a defined expression: ↔ on the same domain
Divide by an expression: ↔ only if it is ≠ 0
Square both sides: ⇒ check candidates

The same operation on both sides preserves equality. The balance represents a bill in hryvnias, not a mass model.

Moving a term abbreviates an operation

In $5x-8=2x+13$ , subtract $2x$ and add 8 on both sides. We obtain $3x=21$ , hence $x=7$ . Changing a sign when moving a term is not magic: it abbreviates adding the opposite expression to both sides.

Adding or subtracting the same defined expression preserves solutions on the original domain. Multiplying or dividing by an expression is equivalent only where it is defined and nonzero. A fixed nonzero number automatically meets this condition. Equality properties, OpenStax §2.2.

Three outcomes for ax = b

After simplification, linear problems reduce to $ax=b$ , with given numbers $a,b$ . If $a\ne0$ , there is one solution: $x=b/a$ .

But $a=0$ needs separate treatment. $0x=0$ is true for every real $x$ . $0x=5$ is false for all $x$ , so no solution exists. For example, $2(x+1)=2x+2$ reduces to $0=0$ , whereas $2(x+1)=2x+3$ reduces to $0=1$ . Dividing zero by zero cannot find $x$ .

Lost and extraneous solutions

The equation $x(x-2)=0$ has solutions 0 and 2. Dividing by $x$ without a condition leaves $x-2=0$ , losing 0: division was forbidden precisely there. Instead, use the zero-product property or treat $x=0$ and $x\ne0$ separately.

Squaring can have the opposite effect. From $x=-2$ we obtain $x^2=4$ , but the second equation also allows $x=2$ . It is a consequence equation: every old solution remains, but new ones can appear. They are extraneous to the original equation. The symbol $\Rightarrow$ marks one direction; $\Leftrightarrow$ marks both.

Branch · VerificationChecking candidates is necessary but not always sufficient.

What substitution cannot repair

Substituting 2 into $x=-2$ rejects the extraneous solution. But if division by $x$ left us only 2 for $x(x-2)=0$ , checking confirms it without revealing the missing 0. Completeness comes from valid transformations or an exhaustive split into cases, not just checking the candidates we found.

Solving an equation means finding all allowed solutions or proving that none exist. In the bill, 70 hryvnias restores the correct total; more complex problems also require us to control every irreversible step.