Pre-Algebra Examples

${(2 x - 4)}^{2} - {(4 x - 6)}^{2} = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x - 4$ and $b = 4 x - 6$ .

$(2 x - 4 + 4 x - 6) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2

Simplify.

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Step 1.2.1

Add $2 x$ and $4 x$ .

$(6 x - 4 - 6) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2.2

Subtract $6$ from $- 4$ .

$(6 x - 10) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2.3

Factor $2$ out of $6 x - 10$ .

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Step 1.2.3.1

Factor $2$ out of $6 x$ .

$(2 (3 x) - 10) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2.3.2

Factor $2$ out of $- 10$ .

$(2 (3 x) + 2 (- 5)) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2.3.3

Factor $2$ out of $2 (3 x) + 2 (- 5)$ .

$2 (3 x - 5) (2 x - 4 - (4 x - 6)) = 0$

Step 1.2.4

Apply the distributive property.

$2 (3 x - 5) (2 x - 4 - (4 x) + 6) = 0$

Step 1.2.5

Multiply $4$ by $- 1$ .

$2 (3 x - 5) (2 x - 4 - 4 x + 6) = 0$

Step 1.2.6

Multiply $- 1$ by $- 6$ .

$2 (3 x - 5) (2 x - 4 - 4 x + 6) = 0$

Step 1.2.7

Subtract $4 x$ from $2 x$ .

$2 (3 x - 5) (- 2 x - 4 + 6) = 0$

Step 1.2.8

Add $- 4$ and $6$ .

$2 (3 x - 5) (- 2 x + 2) = 0$

Step 1.2.9

Factor.

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Step 1.2.9.1

Factor $2$ out of $- 2 x + 2$ .

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Step 1.2.9.1.1

Factor $2$ out of $- 2 x$ .

$2 (3 x - 5) (2 (- x) + 2) = 0$

Step 1.2.9.1.2

Factor $2$ out of $2$ .

$2 (3 x - 5) (2 (- x) + 2 (1)) = 0$

Step 1.2.9.1.3

Factor $2$ out of $2 (- x) + 2 (1)$ .

$2 (3 x - 5) (2 (- x + 1)) = 0$

Step 1.2.9.2

Remove unnecessary parentheses.

$2 (3 x - 5) \cdot (2 (- x + 1)) = 0$

Step 1.2.10

Multiply $2$ by $2$ .

$4 (3 x - 5) (- x + 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 5 = 0$

$- x + 1 = 0$

Step 3

Set $3 x - 5$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Step 3.2

Solve $3 x - 5 = 0$ for $x$ .

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Step 3.2.1

Add $5$ to both sides of the equation.

$3 x = 5$

Step 3.2.2

Divide each term in $3 x = 5$ by $3$ and simplify.

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Step 3.2.2.1

Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Step 3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{3}$

Step 4

Set $- x + 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Step 4.2

Solve $- x + 1 = 0$ for $x$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 4.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 4.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 4.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 5

The final solution is all the values that make $4 (3 x - 5) (- x + 1) = 0$ true.

$x = \frac{5}{3}, 1$