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\[6\div 2(1+2)\]

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 \(1+2\).

\[6\div 2\times 3\]

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 \(6\div 2\).

\[3\times 3\]

3

Simplify.

\[9\]

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.

\[2+x=8\times 3\]

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.

\[x=24-2\]

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.

\[3x=5-7\]

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.

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

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 \({x}^{5}\).

\[{x}^{2+3}{y}^{5+4}\]

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.

\[{({x}^{2})}^{2}-{6}^{2}\]

2

Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).

\[({x}^{2}+6)({x}^{2}-6)\]

Done