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\[2x+5=9\]

1

?

Subtract \(5\) from both sides.

Why did we take this step?

Because we have \(2x+5\) 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.

\[2x=9-5\]

2

Simplify \(9-5\) to \(4\).

\[2x=4\]

3

?

Divide both sides by \(2\).

Why did we take this step?

Because we have \(2x\) on the left side, and we want only \(x\).

Therefore, we

divide

to undo the multiplication.

\[x=\frac{4}{2}\]

4

Simplify \(\frac{4}{2}\) to \(2\).

\[x=2\]

Done

\[3(3-2x)=5(7-5x)\]

1

?

Expand.

Why did we take this step?

Because by expanding, we

distribute the terms and remove the parentheses

, which usually allows us to simplify the expression further.

\[9-6x=35-25x\]

2

?

Subtract \(9\) from both sides.

Why did we take this step?

Because this helps us cancel \(9\). Since our goal is to solve for \(x\),

canceling any term that is not \(x\) is helpful

.

\[-6x=35-25x-9\]

3

Simplify \(35-25x-9\) to \(-25x+26\).

\[-6x=-25x+26\]

4

?

Add \(25x\) to both sides.

Why did we take this step?

Because in the previous step, \(x\) is on both sides of the equation. Since our goal is to solve for \(x\),

we need it on one side only

.

\[-6x+25x=26\]

5

Simplify \(-6x+25x\) to \(19x\).

\[19x=26\]

6

?

Divide both sides by \(19\).

Why did we take this step?

Because we have \(19x\) on the left side, and we want only \(x\).

Therefore, we

divide

to undo the multiplication.

\[x=\frac{26}{19}\]

Done

Decimal Form: 1.368421

\[6x=12\]

1

?

Divide both sides by \(6\).

Why did we take this step?

Because we have \(6x\) on the left side, and we want only \(x\).

Therefore, we

divide

to undo the multiplication.

\[x=\frac{12}{6}\]

2

Simplify \(\frac{12}{6}\) to \(2\).

\[x=2\]

Done

\[\sqrt{x+4}=x+5\]

1

Square both sides.

\[x+4={x}^{2}+10x+25\]

2

Move all terms to one side.

\[x+4-{x}^{2}-10x-25=0\]

3

Simplify \(x+4-{x}^{2}-10x-25\) to \(-9x-21-{x}^{2}\).

\[-9x-21-{x}^{2}=0\]

4

Use the Quadratic Formula.

\[x=\frac{9+\sqrt{3}\imath }{-2},\frac{9-\sqrt{3}\imath }{-2}\]

5

Simplify solutions.

\[x=-\frac{9+\sqrt{3}\imath }{2},-\frac{9-\sqrt{3}\imath }{2}\]

Done

\[{x}^{4}+9{x}^{3}+9{x}^{2}-85x-150\]

1

Factor \({x}^{4}+9{x}^{3}+9{x}^{2}-85x-150\) using Polynomial Division.

\[({x}^{3}+7{x}^{2}-5x-75)(x+2)\]

2

Factor \({x}^{3}+7{x}^{2}-5x-75\) using Polynomial Division.

\[({x}^{2}+10x+25)(x-3)(x+2)\]

3

?

Rewrite \({x}^{2}+10x+25\) in the form \({a}^{2}+2ab+{b}^{2}\), where \(a=x\) and \(b=5\).

Why did we take this step?

Because \({a}^{2}+2ab+{b}^{2}\) is a common expression with a known factored form. This allows us to factor the expression in the next step.

\[({x}^{2}+2(x)(5)+{5}^{2})(x-3)(x+2)\]

4

Use Square of Sum: \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\).

\[{(x+5)}^{2}(x-3)(x+2)\]

Done

\[{w}^{2}+8w-65\]

1

Ask: Which two numbers add up to \(8\) and multiply to \(-65\)?

\(-5\) and \(13\)

2

Rewrite the expression using the above.

\[(w-5)(w+13)\]

Done