Question

How do you use order of operations to simplify `1/7 -: (1/7)^2 - 3/343`?

Answer 1

`2398/343`

Explanation:

Follow the order of operations set out in the acronym PEMDAS

{P-parenthesis (brackets) ,E- exponents (powers), M-multiplication, D-division, A- addition and S- subtraction]

`rArr1/7÷(1/7)^2-3/343`

`=1/7÷(1/7xx1/7)-3/343`

`=1/7÷1/49-3/343larrcolor(red)" bracket/exponent"`

[Change division to multiplication and turn dividing fraction upside down.]

`rArr(1/cancel(7)^1xxcancel(49)^7/1)-3/343larrcolor(red)" division"`

`"Finally " 7-3/343`

Change 7 into a fraction with a denominator of 343

`"that is "7/1xx343/343=2401/343`

`rArr2401/343-3/243=2398/343larrcolor(red)" subtraction"`

Answer 2

`= 6 340/343`

Explanation:

In an expression which has different operations, there is a specific order in which they have to be done.

Identify the number of TERMS first. Each term is treated separately and simplified to a single answer.
These are added or subtracted only in the last line.

Within each term, the order to be followed is:

• brackets,
• followed by powers and roots,
• finally multiplication and division

In this case there are two terms. Simplify each separately.

`=color(blue)(1/7" " div" " (1/7)^2) -color(red)(3/343)" "larr` simplify the brackets
`color(white)(.......................)darr`
`=color(blue)(1/7" " div" " 1/49 -color(red)(3/343)`

`color(white)(.............)darrcolor(white)(.....)darr" "larr`multiply by the reciprocal

`=color(blue)(1/7 " "xx" " 49/1 -color(red)(3/343)`

`=color(blue)(1/cancel7 " "xx" " cancel49^7/1 -color(red)(3/343)`

`=" "color(blue)(7)" " -" "color(red)(3/343)`

`= 6 340/343`