Algebra Examples

$\frac{x - 1}{x + 2} = \frac{10}{3 x - 2}$

Step 1

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.

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

Step 2

Solve the equation for $x$ .

Tap for more steps...

Step 2.1

Simplify $(x - 1) (3 x - 2)$ .

Tap for more steps...

Step 2.1.1

Rewrite.

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

Step 2.1.2

Simplify by adding zeros.

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

Step 2.1.3

Expand $(x - 1) (3 x - 2)$ using the FOIL Method.

Tap for more steps...

Step 2.1.3.1

Apply the distributive property.

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

Step 2.1.3.2

Apply the distributive property.

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

Step 2.1.3.3

Apply the distributive property.

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

Step 2.1.4

Simplify and combine like terms.

Tap for more steps...

Step 2.1.4.1

Simplify each term.

Tap for more steps...

Step 2.1.4.1.1

Rewrite using the commutative property of multiplication.

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

Step 2.1.4.1.2

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

Tap for more steps...

Step 2.1.4.1.2.1

Move $x$ .

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

Step 2.1.4.1.2.2

Multiply $x$ by $x$ .

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

Step 2.1.4.1.3

Move $- 2$ to the left of $x$ .

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

Step 2.1.4.1.4

Multiply $3$ by $- 1$ .

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

Step 2.1.4.1.5

Multiply $- 1$ by $- 2$ .

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

Step 2.1.4.2

Subtract $3 x$ from $- 2 x$ .

$3 x^{2} - 5 x + 2 = (x + 2) \cdot 10$

Step 2.2

Simplify $(x + 2) \cdot 10$ .

Tap for more steps...

Step 2.2.1

Apply the distributive property.

$3 x^{2} - 5 x + 2 = x \cdot 10 + 2 \cdot 10$

Step 2.2.2

Simplify the expression.

Tap for more steps...

Step 2.2.2.1

Move $10$ to the left of $x$ .

$3 x^{2} - 5 x + 2 = 10 \cdot x + 2 \cdot 10$

Step 2.2.2.2

Multiply $2$ by $10$ .

$3 x^{2} - 5 x + 2 = 10 x + 20$

Step 2.3

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

Tap for more steps...

Step 2.3.1

Subtract $10 x$ from both sides of the equation.

$3 x^{2} - 5 x + 2 - 10 x = 20$

Step 2.3.2

Subtract $10 x$ from $- 5 x$ .

$3 x^{2} - 15 x + 2 = 20$

Step 2.4

Subtract $20$ from both sides of the equation.

$3 x^{2} - 15 x + 2 - 20 = 0$

Step 2.5

Subtract $20$ from $2$ .

$3 x^{2} - 15 x - 18 = 0$

Step 2.6

Factor the left side of the equation.

Tap for more steps...

Step 2.6.1

Factor $3$ out of $3 x^{2} - 15 x - 18$ .

Tap for more steps...

Step 2.6.1.1

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) - 15 x - 18 = 0$

Step 2.6.1.2

Factor $3$ out of $- 15 x$ .

$3 (x^{2}) + 3 (- 5 x) - 18 = 0$

Step 2.6.1.3

Factor $3$ out of $- 18$ .

$3 x^{2} + 3 (- 5 x) + 3 \cdot - 6 = 0$

Step 2.6.1.4

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

$3 (x^{2} - 5 x) + 3 \cdot - 6 = 0$

Step 2.6.1.5

Factor $3$ out of $3 (x^{2} - 5 x) + 3 \cdot - 6$ .

$3 (x^{2} - 5 x - 6) = 0$

Step 2.6.2

Factor.

Tap for more steps...

Step 2.6.2.1

Factor $x^{2} - 5 x - 6$ using the AC method.

Tap for more steps...

Step 2.6.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Step 2.6.2.1.2

Write the factored form using these integers.

$3 ((x - 6) (x + 1)) = 0$

Step 2.6.2.2

Remove unnecessary parentheses.

$3 (x - 6) (x + 1) = 0$

Step 2.7

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

$x - 6 = 0$

$x + 1 = 0$

Step 2.8

Set $x - 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.8.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.8.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.9

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

Tap for more steps...

Step 2.9.1

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

$x + 1 = 0$

Step 2.9.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.10

The final solution is all the values that make $3 (x - 6) (x + 1) = 0$ true.

$x = 6, - 1$