Question

WILL MARK BRAINLIEST ASAP!!

Answer 1

$\dfrac{x^2+ax-4}{x+2}+\dfrac{x+b}{x+2}=\dfrac{x^2+ax-4+x+b}{x+2}=\dfrac{x^2+(a+1)x+(b-4)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=x+1\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{(x+1)(x+2)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{(x)(x)+(x)(2)+(1)(x)+(1)(2)}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{x^2+2x+x+2}{x+2}\\\\\dfrac{x^2+(a+1)x+(b-4)}{x+2}=\dfrac{x^2+3x+2}{x+2}\iff a+1=3\ \wedge\ b-4=2\\\\\boxed{a=2\ \wedge\ b=6}$

Answer 2

Answer:

The correct choice for a and b is 2 and 6 respectively.

Step-by-step explanation:

$\frac{x^{2}+ax-4}{x+2}+\frac{x+b}{x+2}$

The least common multiple in the provided expression is x+2. Therefore, the above expression can be written as:

$\frac{x^{2}+ax-4+x+b}{x+2}$

Arrange the like terms together as shown:

$\frac{x^{2}+ax+x-4+b}{x+2}=\frac{x^{2}+x(a+1)+b-4}{x+2}$

It is given that after simplified her answer was x+1. Therefore, this information can be written as:

$\frac{x^{2}+x(a+1)+b-4}{x+2}= x+1$

Multiply numerator and denominator by $x+2$ we get,

$\frac{x^{2}+x(a+1)+b-4}{x+2}= \frac{(x+1)(x+2)}{x+2}$

Further simplify,

$\frac{x^{2}+x(a+1)+b-4}{x+2}= \frac{(x^{2}+2x+x+2)}{x+2}$

Add the like terms:

$\frac{x^{2}+x(a+1)+b-4}{x+2}= \frac{(x^{2}+3x+2)}{x+2}$

Now compare the coefficient of x term and constant term.

a+1=3 and b-4=2

further solve:

a=2 and b=6

Hence, the correct choice for a and b is 2 and 6 respectively.