Question

1.Simplify.

2.Simplify.
3.Which statement best reflects the solution(s) of the equation?

Answer 1

Answer to question 1

We want to simplify

$\frac{\frac{4x}{5+x}}{\frac{6x}{x+2}}$

Let us change the middle bar to a normal division sign by rewriting the expression to obtain,

$\frac{4x}{5+x} \div \frac{6x}{x+2}$

We now multiply the first fraction by the reciprocal of the second fraction to get,

$\frac{4x}{5+x} \times \frac{x+2}{6x}$

We cancel out common factors to obtain,

$\frac{2}{5+x} \times \frac{x+2}{3}$

We multiply out to obtain,

$\frac{2(x+2)}{3(x-5)}$

ANSWER TO QUESTION 2

We want to simplify,

$\frac{\frac{x^2+4x+3}{2x-1}}{\frac{x^2+x}{2x^2-3x+1}}$

Let us change the middle bar to a normal division sign by rewriting the expression to obtain,

$\frac{x^2+4x+3}{2x-1}\div \frac{x^2+x}{2x^2-3x+1}$

We now multiply the first fraction by the reciprocal of the second fraction to get,

$\frac{x^2+4x+3}{2x-1}\times \frac{2x^2-3x+1}{x^2+x}$

We now factor to obtain,

$\frac{(x+1)(x+3)}{2x-1}\times \frac{(x-1)(2x-1)}{x(x+1)}$

We now cancel out common factors to obtain,

$\frac{(x+3)}{1}\times \frac{(x-1)}{x}$

We now multiply out to get,

$\frac{(x-1)(x+3)}{x}$

ANSWER TO QUESTION 3

We want to solve the equation,

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

We need to multiply through by least common multiple of the denominators which is ,

$x(x-1)$

$x(x-1) \times \frac{1}{x-1}+x(x-1) \times \frac{2}{x}=x(x-1) \times \frac{x}{x-1}$

$x+2(x-1)=x(x)$

$x+2x-2=x^2$

$3x-2=x^2$

$x^2-3x+2=0$

$(x-1)(x-2)=0$

$x=1,x=2$

But $x=1$ does not satisfy the equation. It will result in division by zero which is undefined. This is an extraneous solution.

Therefore $x=2$ is the only solution.

The correct answer is D.