\"Know

6\div 2(1+2)

1

?

Simplify

1+2

to

3

.

Why did we take this step?

Because of PEMDAS (the order of operations), we ask the questions below in order.

Any

parentheses

?

Yes.

Any

exponents

? --

Any

multiplication / division

? --

Any

addition / subtraction

? --

Therefore, we

simplify terms in parentheses

first. In other words, we simplify

1+2

.

1

Use a number line.

0

1

2

3

2

Therefore,

1+2=3

.

3

6\div 2\times 3

2

?

Simplify

6\div 2

to

3

.

Why did we take this step?

Because of PEMDAS (the order of operations), we ask the questions below in order.

Any

parentheses

? No.

Any

exponents

? No.

Any

multiplication / division

?

Yes, division.

Any

addition / subtraction

? --

Therefore, we

divide

first. In other words, we simplify

6\div 2

.

1

We can divide

6

items evenly into

3

groups of

2

.

2

Therefore,

6\div 2=3

.

3

3\times 3

3

Simplify.

1

Draw

3

groups of

3

.

2

Therefore,

3\times 3=9

.

9

9

Done

\frac{2+x}{3}=8

1

?

Multiply both sides by

3

.

Why did we take this step?

Because we have

\frac{2+x}{3}

on the left side, and we want only

x

. Using Reverse PEMDAS, we ask the questions below in order.

Any

addition / subtraction

outside parentheses? No.

Any

multiplication / division

outside parentheses?

Yes, division.

Any

exponents

? --

Any

parentheses

? --

Therefore, we

multiply

to undo the division.

1

To remove

3

on the left side, multiply both sides by

3

.

\frac{2+x}{3}\times 3=8\times 3

2

Cancel

3

.

2+x=8\times 3

2+x=8\times 3

2

Simplify

8\times 3

to

24

.

2+x=24

3

?

Subtract

2

from both sides.

Why did we take this step?

Because we have

2+x

on the left side, and we want only

x

.

Therefore, we

subtract

to undo the addition.

1

To remove

2

on the left side, subtract

2

from both sides.

2+x-2=24-2

2

Cancel

2

.

x=24-2

x=24-2

4

Simplify

24-2

to

22

.

x=22

Done

3x+7=5

1

?

Subtract

7

from both sides.

Why did we take this step?

Because we have

3x+7

on the left side, and we want only

x

. Using Reverse PEMDAS, we ask the questions below in order.

Any

addition / subtraction

outside parentheses?

Yes, addition.

Any

multiplication / division

outside parentheses? --

Any

exponents

? --

Any

parentheses

? --

Therefore, we

subtract

to undo the addition.

1

To remove

7

on the left side, subtract

7

from both sides.

3x+7-7=5-7

2

Cancel

7

.

3x=5-7

3x=5-7

2

Simplify

5-7

to

-2

.

3x=-2

3

?

Divide both sides by

3

.

Why did we take this step?

Because we have

3x

on the left side, and we want only

x

.

Therefore, we

divide

to undo the multiplication.

1

To remove

3

on the left side, divide both sides by

3

.

\frac{3x}{3}=-\frac{2}{3}

2

Cancel

3

.

x=-\frac{2}{3}

x=-\frac{2}{3}

Done

Decimal Form: -0.666667

{x}^{2}{x}^{3}{y}^{5}{y}^{4}

1

?

Use Product Rule:

{x}^{a}{x}^{b}={x}^{a+b}

.

Why did we take this step?

Because the

Product Rule

simplifies the expression. Let us take

{x}^{2}{x}^{3}

as an example. You can think of

{x}^{2}

as 2 copies of

x

, and

{x}^{3}

as 3 copies of

x

. Therefore:

In this example, we end up with 5 copies of

x

in total, which is

{x}^{5}

.

1

The Product Rule states that, to multiply two exponents with the same base, we add the exponents while keeping the base. In general:

{x}^{a}{x}^{b}={x}^{a+b}

2

Next, let's apply this rule to

x

in our expression.

{x}^{2}{x}^{3}={x}^{2+3}

3

Next, let's apply this rule to

y

in our expression.

{y}^{5}{y}^{4}={y}^{5+4}

4

Putting this back into the original expression, we have:

{x}^{2+3}{y}^{5+4}

{x}^{2+3}{y}^{5+4}

2

Simplify

2+3

to

5

.

{x}^{5}{y}^{5+4}

3

Simplify

5+4

to

9

.

{x}^{5}{y}^{9}

Done

{x}^{4}-36

1

?

Rewrite it in the form

{a}^{2}-{b}^{2}

, where

a={x}^{2}

and

b=6

.

Why did we take this step?

Because

{a}^{2}-{b}^{2}

is a common expression with a known factored form. This allows us to factor the expression in the next step.

1

Recognize that

{x}^{4}

is a perfect square.

{x}^{4}

=

{x}^{2} \times {x}^{2}

= Perfect Square

2

Recognize that

36

is a perfect square.

36

=

6 \times 6

= Perfect Square

{({x}^{2})}^{2}-{6}^{2}

2

Use Difference of Squares:

{a}^{2}-{b}^{2}=(a+b)(a-b)

.

({x}^{2}+6)({x}^{2}-6)

Done