Example

Simplify [latex]\Large\frac{5-[3+(2\cdot (-6))]}{{{3}^{2}}+2}[/latex]

Answer: This problem has brackets, parentheses, fractions, exponents, multiplication, subtraction, and addition in it. Grouping symbols are handled first. The parentheses around the [latex]-6[/latex] aren’t a grouping symbol; they are simply making it clear that the negative sign belongs to the 6. Start with the innermost set of parentheses that are a grouping symbol. In this example, the innermost set of parentheses would be in the numerator of the fraction, [latex](2\cdot(−6))[/latex]. Begin working out from there. (The fraction line acts as a type of grouping symbol, too; you simplify the numerator and denominator independently, and then divide the numerator by the denominator at the end.)

[latex]\Large\begin{array}{c}\frac{5-\left[3+\left(2\cdot\left(-6\right)\right)\right]}{3^{2}+2}\\\\\frac{5-\left[3+\left(-12\right)\right]}{3^{2}+2}\end{array}[/latex]

Add [latex]3[/latex] and [latex]-12[/latex], which are in brackets, to get [latex]-9[/latex].

[latex]\Large\begin{array}{c}\frac{5-\left[3+\left(-12\right)\right]}{3^{2}+2}\\\\\frac{5-\left[-9\right]}{3^{2}+2}\end{array}[/latex]

Subtract [latex]5–\left[−9\right]=5+9=14[/latex].

[latex]\Large\begin{array}{c}\frac{5-\left[-9\right]}{3^{2}+2}\\\\\frac{14}{3^{2}+2}\end{array}[/latex]

The top of the fraction is all set, but the bottom (denominator) has remained untouched. Apply the order of operations to that as well. Begin by evaluating [latex]3^{2}=9[/latex].

[latex]\Large\begin{array}{c}\frac{14}{3^{2}+2}\\\\\frac{14}{9+2}\end{array}[/latex]

Now add. [latex]9+2=11[/latex].

[latex]\Large\begin{array}{c}\frac{14}{9+2}\\\\\frac{14}{11}\end{array}[/latex]

Answer

[latex-display]\Large\frac{5-\left[3+\left(2\cdot\left(-6\right)\right)\right]}{3^{2}+2}=\frac{14}{11}[/latex-display]

Example

Simplify [latex]\Large\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}[/latex]

Answer: This problem has absolute values, decimals, multiplication, subtraction, and addition in it. Grouping symbols, including absolute value, are handled first. Simplify the numerator, then the denominator. Evaluate [latex]\left|2–6\right|[/latex].

[latex]\Large\begin{array}{c}\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{3+\left|-4\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\end{array}[/latex]

Take the absolute value of [latex]\left|−4\right|[/latex].

[latex]\Large\begin{array}{c}\frac{3+\left|-4\right|}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{3+4}{2\left|3\cdot1.5\right|-\left(-3\right)}\end{array}[/latex]

Add the numbers in the numerator.

[latex]\Large\begin{array}{c}\frac{3+4}{2\left|3\cdot1.5\right|-\left(-3\right)}\\\\\frac{7}{2\left| 3\cdot 1.5 \right|-(-3)}\end{array}[/latex]

Now that the numerator is simplified, turn to the denominator. Evaluate the absolute value expression first. [latex]3 \cdot 1.5 = 4.5[/latex], giving

 [latex]\Large\begin{array}{c}\frac{7}{2\left|{3\cdot{1.5}}\right|-(-3)}\\\\\frac{7}{2\left|{ 4.5}\right|-(-3)}\end{array}[/latex]

The expression “[latex]2\left|4.5\right|[/latex]” reads “[latex]2[/latex] times the absolute value of [latex]4.5[/latex].” Multiply [latex]2[/latex] times [latex]4.5[/latex].

[latex]\Large\begin{array}{c}\frac{7}{2\left|4.5\right|-\left(-3\right)}\\\\\frac{7}{9-\left(-3\right)}\end{array}[/latex]

Subtract.

[latex]\Large\begin{array}{c}\frac{7}{9-\left(-3\right)}\\\\\frac{7}{12}\end{array}[/latex]

Answer

[latex-display]\Large\frac{3+\left|2-6\right|}{2\left|3\cdot1.5\right|-3\left(-3\right)}=\frac{7}{12}[/latex-display]