Subsection A.3 Properties of Equality
Principle A.3.20. Addition Principle of Equality.
If we add (or subtract) the same numerical expression from both sides of an equation, we get an equivalent equation. In symbols, \(a=b\) if and only if \(a+c=b+c\text{.}\) Here, \(a\text{,}\) \(b\text{,}\) and \(c\) may be numbers or algebraic expressions.
Principle A.3.21. Multiplication Principle of Equality.
If we multiply (or divide) both sides of an equation by the same nonzero numerical expression, we get an equivalent equation. In symbols, \(a=b\) if and only if \(a\cdot c=b\cdot c\) whenever \(a\text{,}\) \(b\text{,}\) and \(c\) are numbers or algebraic expressions and \(c\neq 0\text{.}\) Similarly, if \(c\neq 0\text{,}\) then \(a=b\) is equivalent to \(\frac{a}{c}=\frac{b}{c}\text{.}\)
The next principle allows us to switch the left and right sides of any equation.
Principle A.3.22. Symmetry Property of Equality.
The statement \(a=b\) is true if and only if the statement \(b=a\) is true.
In addition, if two expressions are both equal to a common expression, then they must be equal to each other.
Principle A.3.23. Transitive Property of Equality.
If \(a=b\) and \(b=c\) are true statements, then \(a=c\text{.}\)
In this course, we may use this transitivity to equate two different expressions that both give the length of a segment or the area of a figure.
The remaining properties of equality focus on the special role of zero.
Principle A.3.24. Multiplication by Zero.
For any numerical expression \(a\text{,}\) \(a\cdot 0=0\text{.}\)
Principle A.3.25. Zero Product Property.
If \(a\) and \(b\) are any numerical expressions such that \(ab=0\text{,}\) then at least one of the factors \(a\) or \(b\) must equal zero.
Example A.3.26. Solving a Polynomial Equation Using Principles.
Solve \(x^3-5x^2=6x\text{.}\)
Answer.
In symbols, with properties used:
\begin{align*}
x^3-5x^2 \amp = 6x \amp\amp \\
x^3-5x^2-6x \amp = 0 \amp \amp
\text{ (Additive Property of Equality)}\\
x(x^2-5x-6) \amp = 0 \amp\amp \text{ (Factor; Distributive Law)}\\
x(x-6)(x+1) \amp = 0 \amp\amp \text{ (Factor; Distributive Law)}\\
x=0, x=6,\amp \text{ or} x=-1 \amp\amp \text{ (Zero Product Property)}
\end{align*}
Click on solution to see this solution in paragraph form.
Solution.
To solve
\begin{equation*}
x^3-5x^2=6x\text{,}
\end{equation*}
\begin{equation*}
x^3-5x^2-6x=0\text{.}
\end{equation*}
\begin{equation*}
x(x-6)(x+1)=(x^2-6x)(x+1)=x^3-6x^2+x^2-6x=x^3-5x^2-6x\text{.}
\end{equation*}
Substituting into our earlier equation, we see
\begin{equation*}
x(x-6)(x+1)=0\text{.}
\end{equation*}
Therefore,
Zero Product Property informs us that one of the following must be true:
\(x=0\text{,}\) \(x-6=0\text{,}\) or
\(x+1=0\text{.}\) Solving each of these for
\(x\) gives the possible answers,
\(x=0\text{,}\) \(x=6\text{,}\) and
\(x=-1\text{.}\) Substituting each of these into the original equation verifies that the solution consists of these three values,
\(0\text{,}\) \(6\text{,}\) and
\(-1\text{.}\)
Click on answer to see the solution as a series of equations.
