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3 years ago

What is the solution of the following system -2x-y=1 -4x-2y=-1

Mathematics

2 answers:

motikmotik3 years ago

8 0

-2x - y = 1 (1st)

-4x - 2y = -1 (2nd)

Multiply 1st equation by 2

-4x - 2y = 2

Now you have 2 new equations

-4x - 2y = 2

-4x - 2y = -1

----------------------Subtract

0 ≠ 3

Answer: No solutions

andreev551 [17]3 years ago

7 0

Answer: The correct option is

(B) no solution.

Step-by-step explanation: We are given to find the solution to the following system of equations :

$-2x-y=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\-4x-2y=-1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

From equation (i), we have

$-2x-y=1\\\\\Rightarrow y=-2x-1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of y from equation (iii) in equation (ii), we get

$-4x-2(-2x-1)=-1\\\\\Rightarrow -4x+4x+2=-1\\\\\Rightarrow 2=-1,$

which can never be possible.

Therefore, the given system of equations has no solution.

Option (B) is CORRECT.

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Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

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Now , Differentiate both sides with respect to x , we will get

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[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$