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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
outside parentheses?
Any
multiplication / division
outside parentheses? --
Any
exponents
? --
Any
parentheses
? --
Therefore, we
subtract
$2x=9-5$

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

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