We have that
[6x+5]=1+2*(3x+2)
[6x+5]=1+2*3x+2 --------> is not correct ------->1+ 2*(3x+2)=1+2*3x+2*2

then
[6x+5]=1+6x+4---------------> [6x+5]=[6x+5]

this equation is an identity, all real numbers are solutions.

5 0

3 years ago

Read 2 more answers

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

Step-by-step explanation:

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$

6 0

3 years ago