Algebra Examples

$\frac{7}{x + 1} = \frac{2 x + 4}{3 x - 3}$

Step 1

Simplify both sides.

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Step 1.1

Factor $2$ out of $2 x + 4$ .

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Step 1.1.1

Factor $2$ out of $2 x$ .

$\frac{7}{x + 1} = \frac{2 (x) + 4}{3 x - 3}$

Step 1.1.2

Factor $2$ out of $4$ .

$\frac{7}{x + 1} = \frac{2 x + 2 \cdot 2}{3 x - 3}$

Step 1.1.3

Factor $2$ out of $2 x + 2 \cdot 2$ .

$\frac{7}{x + 1} = \frac{2 (x + 2)}{3 x - 3}$

Step 1.2

Factor $3$ out of $3 x - 3$ .

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Step 1.2.1

Factor $3$ out of $3 x$ .

$\frac{7}{x + 1} = \frac{2 (x + 2)}{3 (x) - 3}$

Step 1.2.2

Factor $3$ out of $- 3$ .

$\frac{7}{x + 1} = \frac{2 (x + 2)}{3 (x) + 3 (- 1)}$

Step 1.2.3

Factor $3$ out of $3 (x) + 3 (- 1)$ .

$\frac{7}{x + 1} = \frac{2 (x + 2)}{3 (x - 1)}$

Step 2

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$7 (3 (x - 1)) = (x + 1) (2 (x + 2))$

Step 3

Solve the equation 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.

$(x + 1) \cdot (2 (x + 2)) = 7 \cdot (3 (x - 1))$

Step 3.2

Simplify $(x + 1) \cdot (2 (x + 2))$ .

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Step 3.2.1

Rewrite.

$0 + 0 + (x + 1) \cdot (2 (x + 2)) = 7 \cdot (3 (x - 1))$

Step 3.2.2

Simplify by multiplying through.

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Step 3.2.2.1

Apply the distributive property.

$(x + 1) \cdot (2 x + 2 \cdot 2) = 7 \cdot (3 (x - 1))$

Step 3.2.2.2

Multiply $2$ by $2$ .

$(x + 1) \cdot (2 x + 4) = 7 \cdot (3 (x - 1))$

Step 3.2.3

Expand $(x + 1) (2 x + 4)$ using the FOIL Method.

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Step 3.2.3.1

Apply the distributive property.

$x (2 x + 4) + 1 (2 x + 4) = 7 \cdot (3 (x - 1))$

Step 3.2.3.2

Apply the distributive property.

$x (2 x) + x \cdot 4 + 1 (2 x + 4) = 7 \cdot (3 (x - 1))$

Step 3.2.3.3

Apply the distributive property.

$x (2 x) + x \cdot 4 + 1 (2 x) + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4

Simplify and combine like terms.

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Step 3.2.4.1

Simplify each term.

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Step 3.2.4.1.1

Rewrite using the commutative property of multiplication.

$2 x \cdot x + x \cdot 4 + 1 (2 x) + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.1.2

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

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Step 3.2.4.1.2.1

Move $x$ .

$2 (x \cdot x) + x \cdot 4 + 1 (2 x) + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.1.2.2

Multiply $x$ by $x$ .

$2 x^{2} + x \cdot 4 + 1 (2 x) + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.1.3

Move $4$ to the left of $x$ .

$2 x^{2} + 4 \cdot x + 1 (2 x) + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.1.4

Multiply $2 x$ by $1$ .

$2 x^{2} + 4 x + 2 x + 1 \cdot 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.1.5

Multiply $4$ by $1$ .

$2 x^{2} + 4 x + 2 x + 4 = 7 \cdot (3 (x - 1))$

Step 3.2.4.2

Add $4 x$ and $2 x$ .

$2 x^{2} + 6 x + 4 = 7 \cdot (3 (x - 1))$

Step 3.3

Simplify $7 \cdot (3 (x - 1))$ .

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Step 3.3.1

Apply the distributive property.

$2 x^{2} + 6 x + 4 = 7 \cdot (3 x + 3 \cdot - 1)$

Step 3.3.2

Multiply $3$ by $- 1$ .

$2 x^{2} + 6 x + 4 = 7 \cdot (3 x - 3)$

Step 3.3.3

Apply the distributive property.

$2 x^{2} + 6 x + 4 = 7 (3 x) + 7 \cdot - 3$

Step 3.3.4

Multiply.

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Step 3.3.4.1

Multiply $3$ by $7$ .

$2 x^{2} + 6 x + 4 = 21 x + 7 \cdot - 3$

Step 3.3.4.2

Multiply $7$ by $- 3$ .

$2 x^{2} + 6 x + 4 = 21 x - 21$

Step 3.4

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

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Step 3.4.1

Subtract $21 x$ from both sides of the equation.

$2 x^{2} + 6 x + 4 - 21 x = - 21$

Step 3.4.2

Subtract $21 x$ from $6 x$ .

$2 x^{2} - 15 x + 4 = - 21$

Step 3.5

Add $21$ to both sides of the equation.

$2 x^{2} - 15 x + 4 + 21 = 0$

Step 3.6

Add $4$ and $21$ .

$2 x^{2} - 15 x + 25 = 0$

Step 3.7

Factor by grouping.

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Step 3.7.1

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 $a \cdot c = 2 \cdot 25 = 50$ and whose sum is $b = - 15$ .

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Step 3.7.1.1

Factor $- 15$ out of $- 15 x$ .

$2 x^{2} - 15 x + 25 = 0$

Step 3.7.1.2

Rewrite $- 15$ as $- 5$ plus $- 10$

$2 x^{2} + (- 5 - 10) x + 25 = 0$

Step 3.7.1.3

Apply the distributive property.

$2 x^{2} - 5 x - 10 x + 25 = 0$

Step 3.7.2

Factor out the greatest common factor from each group.

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Step 3.7.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 5 x) - 10 x + 25 = 0$

Step 3.7.2.2

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

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

Step 3.7.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

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

Step 3.8

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

$2 x - 5 = 0$

$x - 5 = 0$

Step 3.9

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 3.9.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 3.9.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 3.9.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 3.9.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 3.9.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 3.9.2.2.2

Simplify the left side.

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Step 3.9.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.9.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 3.9.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 3.10

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

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Step 3.10.1

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

$x - 5 = 0$

Step 3.10.2

Add $5$ to both sides of the equation.

$x = 5$

Step 3.11

The final solution is all the values that make $(2 x - 5) (x - 5) = 0$ true.

$x = \frac{5}{2}, 5$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2}, 5$

Decimal Form:

$x = 2.5, 5$

Mixed Number Form:

$x = 2 \frac{1}{2}, 5$