Pls answer I'll give brainliest to whoever replies first

Mathematics

1 answer:

galben [10]3 years ago

7 0

1: x= 2y - 1, y eR
2: x= 3y + 4, y eR
I think that’s right.-.

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Which product is positive?

svp [43]

Answer and Step-by-step explanation:

The product that would be positive is the last answer choice.

This is because there is 2 negatives.

When two negatives multiply with each other, they become positive.

When three negatives multiply with each other, it becomes negative again.

There is no need to actually solve for the final product, just look for the amount of negatives and remember the properties of multiplying negatives.

#teamtrees #PAW (Plant And Water)

5 0

3 years ago

Someone please help me with q6

Studentka2010 [4]

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\\\\\tan3x=\tan(2x+x)=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}=\dfrac{\tan(x+x)+\tan x}{1-\tan(x+x)\tan x}\\\\=\dfrac{\frac{\tan x+\tan x}{1-\tan x\tan x}+\tan x}{1-\frac{\tan x+\tan x}{1-\tan x\tan x}\tan x}=\dfrac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}$

$=\left(\dfrac{2\tan x}{1-\tan^2x}+\dfrac{\tan x(1-\tan^2x)}{1-\tan^2x}\right):\left(\dfrac{1-\tan^2x}{1-\tan^2x}-\dfrac{2\tan^2x}{1-\tan^2x}\right)\\\\=\dfrac{2\tan x+\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-\tan^2x-2\tan^2x}{1-\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}:\dfrac{1-3\tan^2x}{1-\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-\tan^2x}\cdot\dfrac{1-\tan^2x}{1-3\tan^2x}\\\\=\dfrac{3\tan x-\tan^3x}{1}\cdot\dfrac{1}{1-3\tan^2x}=\dfrac{3\tan x-\tan^3x}{1-3\tan^2x}$