Algebra Examples

$\frac{j^{2} + 2 j}{j^{2} + 9 j + 14} + \frac{1}{j^{2} - 49} = - 10$

Step 1

Factor each term.

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Step 1.1

Factor $j$ out of $j^{2} + 2 j$ .

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Step 1.1.1

Factor $j$ out of $j^{2}$ .

$\frac{j \cdot j + 2 j}{j^{2} + 9 j + 14} + \frac{1}{j^{2} - 49} = - 10$

Step 1.1.2

Factor $j$ out of $2 j$ .

$\frac{j \cdot j + j \cdot 2}{j^{2} + 9 j + 14} + \frac{1}{j^{2} - 49} = - 10$

Step 1.1.3

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

$\frac{j (j + 2)}{j^{2} + 9 j + 14} + \frac{1}{j^{2} - 49} = - 10$

Step 1.2

Factor $j^{2} + 9 j + 14$ using the AC method.

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Step 1.2.1

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 $14$ and whose sum is $9$ .

$2, 7$

Step 1.2.2

Write the factored form using these integers.

$\frac{j (j + 2)}{(j + 2) (j + 7)} + \frac{1}{j^{2} - 49} = - 10$

Step 1.3

Reduce the expression $\frac{j (j + 2)}{(j + 2) (j + 7)}$ by cancelling the common factors.

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Step 1.3.1

Cancel the common factor.

$\frac{j (j + 2)}{(j + 2) (j + 7)} + \frac{1}{j^{2} - 49} = - 10$

Step 1.3.2

Rewrite the expression.

$\frac{j}{j + 7} + \frac{1}{j^{2} - 49} = - 10$

Step 1.4

Rewrite $49$ as $7^{2}$ .

$\frac{j}{j + 7} + \frac{1}{j^{2} - 7^{2}} = - 10$

Step 1.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = j$ and $b = 7$ .

$\frac{j}{j + 7} + \frac{1}{(j + 7) (j - 7)} = - 10$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$j + 7, (j + 7) (j - 7), 1$

Step 2.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.5

The factor for $j + 7$ is $j + 7$ itself.

$(j + 7) = j + 7$

$(j + 7)$ occurs $1$ time.

Step 2.6

The factor for $j - 7$ is $j - 7$ itself.

$(j - 7) = j - 7$

$(j - 7)$ occurs $1$ time.

Step 2.7

The LCM of $j + 7, j + 7, j - 7$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(j + 7) (j - 7)$

Step 3

Multiply each term in $\frac{j}{j + 7} + \frac{1}{(j + 7) (j - 7)} = - 10$ by $(j + 7) (j - 7)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{j}{j + 7} + \frac{1}{(j + 7) (j - 7)} = - 10$ by $(j + 7) (j - 7)$ .

$\frac{j}{j + 7} ((j + 7) (j - 7)) + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $j + 7$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{j}{j + 7} ((j + 7) (j - 7)) + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.1.2

Rewrite the expression.

$j (j - 7) + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.2

Apply the distributive property.

$j \cdot j + j \cdot - 7 + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.3

Multiply $j$ by $j$ .

$j^{2} + j \cdot - 7 + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.4

Move $- 7$ to the left of $j$ .

$j^{2} - 7 \cdot j + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.5

Cancel the common factor of $(j + 7) (j - 7)$ .

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Step 3.2.1.5.1

Cancel the common factor.

$j^{2} - 7 j + \frac{1}{(j + 7) (j - 7)} ((j + 7) (j - 7)) = - 10 ((j + 7) (j - 7))$

Step 3.2.1.5.2

Rewrite the expression.

$j^{2} - 7 j + 1 = - 10 ((j + 7) (j - 7))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Expand $(j + 7) (j - 7)$ using the FOIL Method.

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Step 3.3.1.1

Apply the distributive property.

$j^{2} - 7 j + 1 = - 10 (j (j - 7) + 7 (j - 7))$

Step 3.3.1.2

Apply the distributive property.

$j^{2} - 7 j + 1 = - 10 (j \cdot j + j \cdot - 7 + 7 (j - 7))$

Step 3.3.1.3

Apply the distributive property.

$j^{2} - 7 j + 1 = - 10 (j \cdot j + j \cdot - 7 + 7 j + 7 \cdot - 7)$

Step 3.3.2

Simplify terms.

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Step 3.3.2.1

Combine the opposite terms in $j \cdot j + j \cdot - 7 + 7 j + 7 \cdot - 7$ .

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Step 3.3.2.1.1

Reorder the factors in the terms $j \cdot - 7$ and $7 j$ .

$j^{2} - 7 j + 1 = - 10 (j \cdot j - 7 j + 7 j + 7 \cdot - 7)$

Step 3.3.2.1.2

Add $- 7 j$ and $7 j$ .

$j^{2} - 7 j + 1 = - 10 (j \cdot j + 0 + 7 \cdot - 7)$

Step 3.3.2.1.3

Add $j \cdot j$ and $0$ .

$j^{2} - 7 j + 1 = - 10 (j \cdot j + 7 \cdot - 7)$

Step 3.3.2.2

Simplify each term.

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Step 3.3.2.2.1

Multiply $j$ by $j$ .

$j^{2} - 7 j + 1 = - 10 (j^{2} + 7 \cdot - 7)$

Step 3.3.2.2.2

Multiply $7$ by $- 7$ .

$j^{2} - 7 j + 1 = - 10 (j^{2} - 49)$

Step 3.3.2.3

Simplify by multiplying through.

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Step 3.3.2.3.1

Apply the distributive property.

$j^{2} - 7 j + 1 = - 10 j^{2} - 10 \cdot - 49$

Step 3.3.2.3.2

Multiply $- 10$ by $- 49$ .

$j^{2} - 7 j + 1 = - 10 j^{2} + 490$

Step 4

Solve the equation.

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Step 4.1

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

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Step 4.1.1

Add $10 j^{2}$ to both sides of the equation.

$j^{2} - 7 j + 1 + 10 j^{2} = 490$

Step 4.1.2

Add $j^{2}$ and $10 j^{2}$ .

$11 j^{2} - 7 j + 1 = 490$

Step 4.2

Move all terms to the left side of the equation and simplify.

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Step 4.2.1

Subtract $490$ from both sides of the equation.

$11 j^{2} - 7 j + 1 - 490 = 0$

Step 4.2.2

Subtract $490$ from $1$ .

$11 j^{2} - 7 j - 489 = 0$

Step 4.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.4

Substitute the values $a = 11$ , $b = - 7$ , and $c = - 489$ into the quadratic formula and solve for $j$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (11 \cdot - 489)}}{2 \cdot 11}$

Step 4.5

Simplify.

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Step 4.5.1

Simplify the numerator.

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Step 4.5.1.1

Raise $- 7$ to the power of $2$ .

$j = \frac{7 \pm \sqrt{49 - 4 \cdot 11 \cdot - 489}}{2 \cdot 11}$

Step 4.5.1.2

Multiply $- 4 \cdot 11 \cdot - 489$ .

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Step 4.5.1.2.1

Multiply $- 4$ by $11$ .

$j = \frac{7 \pm \sqrt{49 - 44 \cdot - 489}}{2 \cdot 11}$

Step 4.5.1.2.2

Multiply $- 44$ by $- 489$ .

$j = \frac{7 \pm \sqrt{49 + 21516}}{2 \cdot 11}$

Step 4.5.1.3

Add $49$ and $21516$ .

$j = \frac{7 \pm \sqrt{21565}}{2 \cdot 11}$

Step 4.5.2

Multiply $2$ by $11$ .

$j = \frac{7 \pm \sqrt{21565}}{22}$

Step 4.6

The final answer is the combination of both solutions.

$j = \frac{7 + \sqrt{21565}}{22}, \frac{7 - \sqrt{21565}}{22}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$j = \frac{7 + \sqrt{21565}}{22}, \frac{7 - \sqrt{21565}}{22}$

Decimal Form:

$j = 6.99319381 \dots, - 6.35683017 \dots$