Solve for a limit as x approaches -4 of (4x^2+4x-8)/(ax^2+bx+8)=9/5

$lim_{x \to - 4} \frac{4 x^{2} + 4 x - 8}{a x^{2} + b x + 8} = \frac{9}{5}$

Step 1

Simplify both sides.

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 4$ .

$\frac{lim_{x \to - 4} 4 x^{2} + 4 x - 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 4$ .

$\frac{lim_{x \to - 4} 4 x^{2} + lim_{x \to - 4} 4 x - lim_{x \to - 4} 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 lim_{x \to - 4} x^{2} + lim_{x \to - 4} 4 x - lim_{x \to - 4} 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{4 {(lim_{x \to - 4} x)}^{2} + lim_{x \to - 4} 4 x - lim_{x \to - 4} 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - lim_{x \to - 4} 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Evaluate the limit of $8$ which is constant as $x$ approaches $- 4$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{lim_{x \to - 4} a x^{2} + b x + 8} = \frac{9}{5}$

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 4$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{lim_{x \to - 4} a x^{2} + lim_{x \to - 4} b x + lim_{x \to - 4} 8} = \frac{9}{5}$

Move the term $a$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{a lim_{x \to - 4} x^{2} + lim_{x \to - 4} b x + lim_{x \to - 4} 8} = \frac{9}{5}$

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{a {(lim_{x \to - 4} x)}^{2} + lim_{x \to - 4} b x + lim_{x \to - 4} 8} = \frac{9}{5}$

Move the term $b$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{a {(lim_{x \to - 4} x)}^{2} + b lim_{x \to - 4} x + lim_{x \to - 4} 8} = \frac{9}{5}$

Evaluate the limit of $8$ which is constant as $x$ approaches $- 4$ .

$\frac{4 {(lim_{x \to - 4} x)}^{2} + 4 lim_{x \to - 4} x - 1 \cdot 8}{a {(lim_{x \to - 4} x)}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Simplify the numerator.

Let $u = lim_{x \to - 4} x$ . Substitute $u$ for all occurrences of $lim_{x \to - 4} x$ .

$\frac{4 u^{2} + 4 u - 8}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $4$ out of $4 u^{2} + 4 u - 8$ .

Factor $4$ out of $4 u^{2}$ .

$\frac{4 (u^{2}) + 4 u - 8}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $4$ out of $4 u$ .

$\frac{4 (u^{2}) + 4 (u) - 8}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $4$ out of $- 8$ .

$\frac{4 (u^{2}) + 4 u + 4 \cdot - 2}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $4$ out of $4 (u^{2}) + 4 u$ .

$\frac{4 (u^{2} + u) + 4 \cdot - 2}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $4$ out of $4 (u^{2} + u) + 4 \cdot - 2$ .

$\frac{4 (u^{2} + u - 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Factor $u^{2} + u - 2$ using the AC method.

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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$\frac{4 ((u - 1) (u + 2))}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

$\frac{4 (u - 1) (u + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Replace all occurrences of $u$ with $lim_{x \to - 4} x$ .

$\frac{4 (lim_{x \to - 4} x - 1) (lim_{x \to - 4} x + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Simplify.

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{4 (- 4 - 1) (lim_{x \to - 4} x + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Subtract $1$ from $- 4$ .

$\frac{4 \cdot - 5 (lim_{x \to - 4} x + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Multiply $4$ by $- 5$ .

$\frac{- 20 (lim_{x \to - 4} x + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{- 20 (- 4 + 2)}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Add $- 4$ and $2$ .

$\frac{- 20 \cdot - 2}{a {lim_{x \to - 4} x}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Simplify the denominator.

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{- 20 \cdot - 2}{a {(- 4)}^{2} + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Raise $- 4$ to the power of $2$ .

$\frac{- 20 \cdot - 2}{a \cdot 16 + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Move $16$ to the left of $a$ .

$\frac{- 20 \cdot - 2}{16 \cdot a + b lim_{x \to - 4} x + 8} = \frac{9}{5}$

Evaluate the limit of $x$ by plugging in $- 4$ for $x$ .

$\frac{- 20 \cdot - 2}{16 a + b \cdot - 4 + 8} = \frac{9}{5}$

Move $- 4$ to the left of $b$ .

$\frac{- 20 \cdot - 2}{16 a - 4 \cdot b + 8} = \frac{9}{5}$

Factor $4$ out of $16 a - 4 b + 8$ .

Factor $4$ out of $16 a$ .

$\frac{- 20 \cdot - 2}{4 (4 a) - 4 b + 8} = \frac{9}{5}$

Factor $4$ out of $- 4 b$ .

$\frac{- 20 \cdot - 2}{4 (4 a) + 4 (- b) + 8} = \frac{9}{5}$

Factor $4$ out of $8$ .

$\frac{- 20 \cdot - 2}{4 (4 a) + 4 (- b) + 4 (2)} = \frac{9}{5}$

Factor $4$ out of $4 (4 a) + 4 (- b)$ .

$\frac{- 20 \cdot - 2}{4 (4 a - b) + 4 (2)} = \frac{9}{5}$

Factor $4$ out of $4 (4 a - b) + 4 (2)$ .

$\frac{- 20 \cdot - 2}{4 (4 a - b + 2)} = \frac{9}{5}$

Multiply $- 20$ by $- 2$ .

$\frac{40}{4 (4 a - b + 2)} = \frac{9}{5}$

Cancel the common factor of $40$ and $4$ .

Factor $4$ out of $40$ .

$\frac{4 \cdot 10}{4 (4 a - b + 2)} = \frac{9}{5}$

Cancel the common factors.

Cancel the common factor.

$\frac{4 \cdot 10}{4 (4 a - b + 2)} = \frac{9}{5}$

Rewrite the expression.

$\frac{10}{4 a - b + 2} = \frac{9}{5}$

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.

$10 \cdot 5 = (4 a - b + 2) \cdot 9$

Step 3

Solve the equation for $a$ .

Rewrite the equation as $(4 a - b + 2) \cdot 9 = 10 \cdot 5$ .

$(4 a - b + 2) \cdot 9 = 10 \cdot 5$

Multiply $10$ by $5$ .

$(4 a - b + 2) \cdot 9 = 50$

Divide each term in $((4 a - b + 2) \cdot 9) = 50$ by $9$ and simplify.

Divide each term in $((4 a - b + 2) \cdot 9) = 50$ by $9$ .

$\frac{((4 a - b + 2) \cdot 9)}{9} = \frac{50}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{(4 a - b + 2) \cdot 9}{9} = \frac{50}{9}$

Divide $4 a - b + 2$ by $1$ .

$4 a - b + 2 = \frac{50}{9}$

Move all terms not containing $a$ to the right side of the equation.

Add $b$ to both sides of the equation.

$4 a + 2 = \frac{50}{9} + b$

Subtract $2$ from both sides of the equation.

$4 a = \frac{50}{9} + b - 2$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$4 a = b + \frac{50}{9} - 2 \cdot \frac{9}{9}$

Combine $- 2$ and $\frac{9}{9}$ .

$4 a = b + \frac{50}{9} + \frac{- 2 \cdot 9}{9}$

Combine the numerators over the common denominator.

$4 a = b + \frac{50 - 2 \cdot 9}{9}$

Simplify the numerator.

Multiply $- 2$ by $9$ .

$4 a = b + \frac{50 - 18}{9}$

Subtract $18$ from $50$ .

$4 a = b + \frac{32}{9}$

Divide each term in $4 a = b + \frac{32}{9}$ by $4$ and simplify.

Divide each term in $4 a = b + \frac{32}{9}$ by $4$ .

$\frac{4 a}{4} = \frac{b}{4} + \frac{\frac{32}{9}}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} = \frac{b}{4} + \frac{\frac{32}{9}}{4}$

Divide $a$ by $1$ .

$a = \frac{b}{4} + \frac{\frac{32}{9}}{4}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$a = \frac{b}{4} + \frac{32}{9} \cdot \frac{1}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $32$ .

$a = \frac{b}{4} + \frac{4 (8)}{9} \cdot \frac{1}{4}$

Cancel the common factor.

$a = \frac{b}{4} + \frac{4 \cdot 8}{9} \cdot \frac{1}{4}$

Rewrite the expression.

$a = \frac{b}{4} + \frac{8}{9}$

Solve for a limit as x approaches -4 of (4x^2+4x-8)/(ax^2+bx+8)=9/5

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