Algebra Examples

$\frac{4 x^{2} - 1}{3} = x (10 x - 9)$

Step 1

Multiply both sides by $3$ .

$\frac{4 x^{2} - 1}{3} \cdot 3 = x (10 x - 9) \cdot 3$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $\frac{4 x^{2} - 1}{3} \cdot 3$ .

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Step 2.1.1.1

Simplify the numerator.

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Step 2.1.1.1.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{{(2 x)}^{2} - 1}{3} \cdot 3 = x (10 x - 9) \cdot 3$

Step 2.1.1.1.2

Rewrite $1$ as $1^{2}$ .

$\frac{{(2 x)}^{2} - 1^{2}}{3} \cdot 3 = x (10 x - 9) \cdot 3$

Step 2.1.1.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$\frac{(2 x + 1) (2 x - 1)}{3} \cdot 3 = x (10 x - 9) \cdot 3$

Step 2.1.1.2

Cancel the common factor of $3$ .

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Step 2.1.1.2.1

Cancel the common factor.

$\frac{(2 x + 1) (2 x - 1)}{3} \cdot 3 = x (10 x - 9) \cdot 3$

Step 2.1.1.2.2

Rewrite the expression.

$(2 x + 1) (2 x - 1) = x (10 x - 9) \cdot 3$

Step 2.1.1.3

Expand $(2 x + 1) (2 x - 1)$ using the FOIL Method.

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Step 2.1.1.3.1

Apply the distributive property.

$2 x (2 x - 1) + 1 (2 x - 1) = x (10 x - 9) \cdot 3$

Step 2.1.1.3.2

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 + 1 (2 x - 1) = x (10 x - 9) \cdot 3$

Step 2.1.1.3.3

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4

Simplify terms.

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Step 2.1.1.4.1

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

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Step 2.1.1.4.1.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

$2 x (2 x) - 1 \cdot 2 x + 1 \cdot 2 x + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.1.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$2 x (2 x) + 0 + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.1.3

Add $2 x (2 x)$ and $0$ .

$2 x (2 x) + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.2

Simplify each term.

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Step 2.1.1.4.2.1

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.2.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.1.4.2.2.1

Move $x$ .

$2 \cdot 2 (x \cdot x) + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.2.2.2

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.2.3

Multiply $2$ by $2$ .

$4 x^{2} + 1 \cdot - 1 = x (10 x - 9) \cdot 3$

Step 2.1.1.4.2.4

Multiply $- 1$ by $1$ .

$4 x^{2} - 1 = x (10 x - 9) \cdot 3$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $x (10 x - 9) \cdot 3$ .

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Step 2.2.1.1

Simplify by multiplying through.

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Step 2.2.1.1.1

Apply the distributive property.

$4 x^{2} - 1 = (x (10 x) + x \cdot - 9) \cdot 3$

Step 2.2.1.1.2

Reorder.

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Step 2.2.1.1.2.1

Rewrite using the commutative property of multiplication.

$4 x^{2} - 1 = (10 x \cdot x + x \cdot - 9) \cdot 3$

Step 2.2.1.1.2.2

Move $- 9$ to the left of $x$ .

$4 x^{2} - 1 = (10 x \cdot x - 9 \cdot x) \cdot 3$

Step 2.2.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.1.2.1

Move $x$ .

$4 x^{2} - 1 = (10 (x \cdot x) - 9 \cdot x) \cdot 3$

Step 2.2.1.2.2

Multiply $x$ by $x$ .

$4 x^{2} - 1 = (10 x^{2} - 9 \cdot x) \cdot 3$

$4 x^{2} - 1 = (10 x^{2} - 9 x) \cdot 3$

Step 2.2.1.3

Simplify by multiplying through.

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Step 2.2.1.3.1

Apply the distributive property.

$4 x^{2} - 1 = 10 x^{2} \cdot 3 - 9 x \cdot 3$

Step 2.2.1.3.2

Multiply.

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Step 2.2.1.3.2.1

Multiply $3$ by $10$ .

$4 x^{2} - 1 = 30 x^{2} - 9 x \cdot 3$

Step 2.2.1.3.2.2

Multiply $3$ by $- 9$ .

$4 x^{2} - 1 = 30 x^{2} - 27 x$

Step 3

Solve for $x$ .

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Step 3.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$30 x^{2} - 27 x = 4 x^{2} - 1$

Step 3.2

Move all terms containing $x$ to the left side of the equation.

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Step 3.2.1

Subtract $4 x^{2}$ from both sides of the equation.

$30 x^{2} - 27 x - 4 x^{2} = - 1$

Step 3.2.2

Subtract $4 x^{2}$ from $30 x^{2}$ .

$26 x^{2} - 27 x = - 1$

Step 3.3

Add $1$ to both sides of the equation.

$26 x^{2} - 27 x + 1 = 0$

Step 3.4

Factor by grouping.

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Step 3.4.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 26 \cdot 1 = 26$ and whose sum is $b = - 27$ .

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Step 3.4.1.1

Factor $- 27$ out of $- 27 x$ .

$26 x^{2} - 27 x + 1 = 0$

Step 3.4.1.2

Rewrite $- 27$ as $- 1$ plus $- 26$

$26 x^{2} + (- 1 - 26) x + 1 = 0$

Step 3.4.1.3

Apply the distributive property.

$26 x^{2} - 1 x - 26 x + 1 = 0$

Step 3.4.2

Factor out the greatest common factor from each group.

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Step 3.4.2.1

Group the first two terms and the last two terms.

$(26 x^{2} - 1 x) - 26 x + 1 = 0$

Step 3.4.2.2

Factor out the greatest common factor (GCF) from each group.

$x (26 x - 1) - (26 x - 1) = 0$

Step 3.4.3

Factor the polynomial by factoring out the greatest common factor, $26 x - 1$ .

$(26 x - 1) (x - 1) = 0$

Step 3.5

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

$26 x - 1 = 0$

$x - 1 = 0$

Step 3.6

Set $26 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $26 x - 1$ equal to $0$ .

$26 x - 1 = 0$

Step 3.6.2

Solve $26 x - 1 = 0$ for $x$ .

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Step 3.6.2.1

Add $1$ to both sides of the equation.

$26 x = 1$

Step 3.6.2.2

Divide each term in $26 x = 1$ by $26$ and simplify.

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Step 3.6.2.2.1

Divide each term in $26 x = 1$ by $26$ .

$\frac{26 x}{26} = \frac{1}{26}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $26$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{26 x}{26} = \frac{1}{26}$

Step 3.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{26}$

Step 3.7

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

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Step 3.7.1

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

$x - 1 = 0$

Step 3.7.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.8

The final solution is all the values that make $(26 x - 1) (x - 1) = 0$ true.

$x = \frac{1}{26}, 1$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{26}, 1$

Decimal Form:

$x = 0.0\overline{384615}, 1$