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3 years ago

Is (6, –2) a solution to this system of equations? y = –1/6 x − 1 y = 1/6 x − 3

Mathematics

1 answer:

Liula [17]3 years ago

5 0

Answer:

Yes, (6, -2) is a solution to the given system of equations.

Step-by-step explanation:

Please write y = –1/6 x − 1 y = 1/6 x − 3 as follows, for greater clarity:

y = (–1/6)x − 1

y = (1/6)x − 3

Let's actually solve this system:

y = (–1/6)x − 1

y = (1/6)x − 3

-----------------------

2y = -4, or y = -2

Now find x. Arbitrarily we choose to use the first equation for this purpose:

y = (-1/6)x - 1. We set y = -2 and find x: -2 = (-1/6)x - 1

Combining the constants, we get -1 = (-1/6)x, or 6 = x

Yes, (6, -2) is a solution to the given system of equations.

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3 years ago

1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β

slava [35]

Step-by-step explanation:

$\tan \beta = \frac{1}{ \cot \beta } = - \frac{3}{2 \sqrt{10} } = - \frac{3 \sqrt{10} }{20}$

$\csc \beta \tan \beta = \frac{1}{ \cos \beta } = \sec \beta$

Therefore,

$\sec \beta = ( \frac{7}{3} )( - \frac{3 \sqrt{10} }{20} ) = - \frac{7 \sqrt{10} }{20}$

$\csc y = \frac{1}{ \sin y} = - \frac{ \sqrt{6} }{2}$

$= > \sin y = - \frac{ \sqrt{6} }{3}$

Use the identity

$\cos y = \sqrt{1 - \sin ^{2} y} \: \: \: \: \: \: \: \: \: \: \: \\ = \sqrt{1 - {( - \frac{ \sqrt{6} }{3}) }^{2} } = - \frac{ \sqrt{3} }{3}$

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

$\tan \beta = \frac{ \sin y}{ \cos y} =( - \frac{ \sqrt{6} }{3} )( - \frac{3}{ \sqrt{3} } ) = \sqrt{2}$

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

$\sin x = \sqrt{1 - \cos ^{2} x} = \sqrt{ \frac{11}{6} }$

Solving for tan x:

$\tan x = \frac{ \sin x}{ \cos x} = (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) = \frac{ \sqrt{66} }{5}$

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