Math / Algebra

QuestionFind the partial fraction decomposition for the rational expression. $\begin{array}{l} \frac{9 x+39}{(x+3)(x+7)} \\ \frac{9 x+39}{(x+3)(x+7)}= \end{array}$ $\square$

Studdy Solution

STEP 1

What is this asking? We're turning one big fraction into the sum of two smaller, simpler fractions! Watch out! Don't mix up the numerators when we split the fraction.
Keep those variables straight!

STEP 2

1. Set up the partial fraction decomposition.
2. Multiply and group like terms.
3. Solve for the unknowns.
4. Write the decomposition.

STEP 3

Alright, let's break down this fraction!
We're aiming to rewrite our big fraction as the sum of two smaller ones.
We can write this as: $\frac{9x + 39}{(x+3)(x+7)} = \frac{A}{x+3} + \frac{B}{x+7}$ where $A$ and $B$ are the mystery numerators we need to find!

STEP 4

To solve for $A$ and $B$ , let's multiply both sides of our equation by $(x+3)(x+7)$ .
This is like giving both sides the same superpower, so they stay balanced!
This gives us: $9x + 39 = A(x+7) + B(x+3)$

STEP 5

Now, let's expand things out: $9x + 39 = Ax + 7A + Bx + 3B$

STEP 6

Let's group the terms with $x$ together and the constant terms together: $9x + 39 = (A+B)x + (7A + 3B)$

STEP 7

Now, the magic happens!
For these two sides to be equal, the coefficients of $x$ must be equal, and the constant terms must be equal.
This gives us a system of equations: $A + B = 9$ $7A + 3B = 39$

STEP 8

Let's solve for $A$ and $B$ !
Multiply the first equation by $-3$ to eliminate $B$ : $-3A - 3B = -27$ $7A + 3B = 39$

STEP 9

Now, add the two equations together: $4A = 12$ So, $A = \frac{12}{4} = \textbf{3}$ !

STEP 10

Substitute $A = \textbf{3}$ back into the equation $A + B = 9$ : $\textbf{3} + B = 9$ $B = 9 - \textbf{3} = \textbf{6}$ We found them! $A = \textbf{3}$ and $B = \textbf{6}$ !

STEP 11

Finally, let's plug our values for $A$ and $B$ back into our initial setup: $\frac{9x + 39}{(x+3)(x+7)} = \frac{\textbf{3}}{x+3} + \frac{\textbf{6}}{x+7}$ Boom! We did it!

STEP 12

$\frac{9x + 39}{(x+3)(x+7)} = \frac{3}{x+3} + \frac{6}{x+7}$

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QuestionFind the partial fraction decomposition for the rational expression. $\begin{array}{l} \frac{9 x+39}{(x+3)(x+7)} \\ \frac{9 x+39}{(x+3)(x+7)}= \end{array}$ $\square$

Studdy Solution

STEP 1

What is this asking? We're turning one big fraction into the sum of two smaller, simpler fractions! Watch out! Don't mix up the numerators when we split the fraction.
Keep those variables straight!

STEP 2

1. Set up the partial fraction decomposition.
2. Multiply and group like terms.
3. Solve for the unknowns.
4. Write the decomposition.

STEP 3

Alright, let's break down this fraction!
We're aiming to rewrite our big fraction as the sum of two smaller ones.
We can write this as: $\frac{9x + 39}{(x+3)(x+7)} = \frac{A}{x+3} + \frac{B}{x+7}$ where $A$ and $B$ are the mystery numerators we need to find!

STEP 4

To solve for $A$ and $B$ , let's multiply both sides of our equation by $(x+3)(x+7)$ .
This is like giving both sides the same superpower, so they stay balanced!
This gives us: $9x + 39 = A(x+7) + B(x+3)$

STEP 5

Now, let's expand things out: $9x + 39 = Ax + 7A + Bx + 3B$

STEP 6

Let's group the terms with $x$ together and the constant terms together: $9x + 39 = (A+B)x + (7A + 3B)$

STEP 7

Now, the magic happens!
For these two sides to be equal, the coefficients of $x$ must be equal, and the constant terms must be equal.
This gives us a system of equations: $A + B = 9$ $7A + 3B = 39$

STEP 8

Let's solve for $A$ and $B$ !
Multiply the first equation by $-3$ to eliminate $B$ : $-3A - 3B = -27$ $7A + 3B = 39$

STEP 9

Now, add the two equations together: $4A = 12$ So, $A = \frac{12}{4} = \textbf{3}$ !

STEP 10

Substitute $A = \textbf{3}$ back into the equation $A + B = 9$ : $\textbf{3} + B = 9$ $B = 9 - \textbf{3} = \textbf{6}$ We found them! $A = \textbf{3}$ and $B = \textbf{6}$ !

STEP 11

Finally, let's plug our values for $A$ and $B$ back into our initial setup: $\frac{9x + 39}{(x+3)(x+7)} = \frac{\textbf{3}}{x+3} + \frac{\textbf{6}}{x+7}$ Boom! We did it!

STEP 12

$\frac{9x + 39}{(x+3)(x+7)} = \frac{3}{x+3} + \frac{6}{x+7}$

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QuestionFind the partial fraction decomposition for the rational expression. 9x+39(x+3)(x+7)9x+39(x+3)(x+7)=\begin{array}{l} \frac{9 x+39}{(x+3)(x+7)} \\ \frac{9 x+39}{(x+3)(x+7)}= \end{array}(x+3)(x+7)9x+39​(x+3)(x+7)9x+39​=​ □\square□

Studdy Solution

STEP 1

What is this asking? We're turning one big fraction into the sum of two smaller, simpler fractions! Watch out! Don't mix up the numerators when we split the fraction.Keep those variables straight!

STEP 2

1. Set up the partial fraction decomposition.2. Multiply and group like terms.3. Solve for the unknowns.4. Write the decomposition.

STEP 3

STEP 4

To solve for AAA and BBB, let's **multiply both sides** of our equation by (x+3)(x+7)(x+3)(x+7)(x+3)(x+7).This is like giving both sides the same superpower, so they stay balanced!This gives us: 9x+39=A(x+7)+B(x+3) 9x + 39 = A(x+7) + B(x+3) 9x+39=A(x+7)+B(x+3)

STEP 5

Now, let's **expand** things out: 9x+39=Ax+7A+Bx+3B 9x + 39 = Ax + 7A + Bx + 3B 9x+39=Ax+7A+Bx+3B

STEP 6

Let's **group** the terms with xxx together and the **constant terms** together: 9x+39=(A+B)x+(7A+3B) 9x + 39 = (A+B)x + (7A + 3B) 9x+39=(A+B)x+(7A+3B)

STEP 7

Now, the **magic happens**!For these two sides to be equal, the coefficients of xxx must be equal, and the constant terms must be equal.This gives us a system of equations: A+B=9 A + B = 9 A+B=9 7A+3B=39 7A + 3B = 39 7A+3B=39

STEP 8

Let's **solve** for AAA and BBB!Multiply the first equation by −3-3−3 to **eliminate** BBB: −3A−3B=−27 -3A - 3B = -27 −3A−3B=−27 7A+3B=39 7A + 3B = 39 7A+3B=39

STEP 9

Now, **add** the two equations together: 4A=12 4A = 12 4A=12 So, A=124=3A = \frac{12}{4} = \textbf{3}A=412​=3!

STEP 10

Substitute A=3A = \textbf{3}A=3 back into the equation A+B=9A + B = 9A+B=9: 3+B=9 \textbf{3} + B = 9 3+B=9 B=9−3=6 B = 9 - \textbf{3} = \textbf{6} B=9−3=6We found them! A=3A = \textbf{3}A=3 and B=6B = \textbf{6}B=6!

STEP 11

Finally, let's **plug** our values for AAA and BBB back into our **initial setup**: 9x+39(x+3)(x+7)=3x+3+6x+7 \frac{9x + 39}{(x+3)(x+7)} = \frac{\textbf{3}}{x+3} + \frac{\textbf{6}}{x+7} (x+3)(x+7)9x+39​=x+33​+x+76​ Boom! We did it!

STEP 12

9x+39(x+3)(x+7)=3x+3+6x+7 \frac{9x + 39}{(x+3)(x+7)} = \frac{3}{x+3} + \frac{6}{x+7} (x+3)(x+7)9x+39​=x+33​+x+76​

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QuestionFind the partial fraction decomposition for the rational expression. $\begin{array}{l} \frac{9 x+39}{(x+3)(x+7)} \\ \frac{9 x+39}{(x+3)(x+7)}= \end{array}$ $\square$

What is this asking? We're turning one big fraction into the sum of two smaller, simpler fractions! Watch out! Don't mix up the numerators when we split the fraction.
Keep those variables straight!

1. Set up the partial fraction decomposition.
2. Multiply and group like terms.
3. Solve for the unknowns.
4. Write the decomposition.

To solve for $A$ and $B$ , let's multiply both sides of our equation by $(x+3)(x+7)$ .
This is like giving both sides the same superpower, so they stay balanced!
This gives us: $9x + 39 = A(x+7) + B(x+3)$

Now, let's expand things out: $9x + 39 = Ax + 7A + Bx + 3B$

Let's group the terms with $x$ together and the constant terms together: $9x + 39 = (A+B)x + (7A + 3B)$

Now, the magic happens!
For these two sides to be equal, the coefficients of $x$ must be equal, and the constant terms must be equal.
This gives us a system of equations: $A + B = 9$ $7A + 3B = 39$

Let's solve for $A$ and $B$ !
Multiply the first equation by $-3$ to eliminate $B$ : $-3A - 3B = -27$ $7A + 3B = 39$

Now, add the two equations together: $4A = 12$ So, $A = \frac{12}{4} = \textbf{3}$ !

Substitute $A = \textbf{3}$ back into the equation $A + B = 9$ : $\textbf{3} + B = 9$ $B = 9 - \textbf{3} = \textbf{6}$ We found them! $A = \textbf{3}$ and $B = \textbf{6}$ !

Finally, let's plug our values for $A$ and $B$ back into our initial setup: $\frac{9x + 39}{(x+3)(x+7)} = \frac{\textbf{3}}{x+3} + \frac{\textbf{6}}{x+7}$ Boom! We did it!

$\frac{9x + 39}{(x+3)(x+7)} = \frac{3}{x+3} + \frac{6}{x+7}$