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Mariana [72]
3 years ago
5

Solve each system of equations by substitution X + 3y = 6 -x + y = -7

Mathematics
1 answer:
notka56 [123]3 years ago
4 0

Answer:

x = 6.75

y = -0.25

Step-by-step explanation:

Given :

x + 3y = 6 ..................... equation 1

-x + y = -7 ................... equation 2

solving the system of equation by substitution method , from equation 1 , make x the subject of the formula , that is

x = 6 - 3y ........................... equation 3

substitute equation 3 into equation 2 , that is

- ( 6 - 3y) + y = -7

-6 + 3y + y = - 7

-6 + 4y = -7

4y = -7 + 6

4y = -1

y = \frac{-1}{4} = -0.25

substitute  y = \frac{-1}{4} into equation 3 to find the value of x , that is

x = 6 - 3 (\frac{-1}{4})

x = 6 + \frac{3}{4}

x = \frac{27}{4}

x = 6. 75

Therefore :

x = 6. 75

y = - 0.25

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If f(x) = 6x-11, what is f(-6)?​
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If f(x) = (6x-11), then f(-6) = -47.

We are provided in the question statement with a function "f(x)" whose output is a polynomial of 1 variable and degree 1.

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\rm\displaystyle  0,\pm\pi

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =  \tan( \alpha )  \tan( \beta )  \tan( \gamma )  -  \tan( \gamma )

factor out tanγ:

\rm\displaystyle  \tan( \alpha )  +  \tan( \beta ) =   \tan( \gamma ) (\tan( \alpha )  \tan( \beta ) -  1)

divide both sides by tanαtanβ-1 and that yields:

\rm\displaystyle   \tan( \gamma ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

multiply both numerator and denominator by-1 which yields:

\rm\displaystyle   \tan( \gamma ) =   -  \bigg(\frac{ \tan( \alpha )  +  \tan( \beta ) }{ 1 - \tan( \alpha )  \tan( \beta )   } \bigg)

recall angle sum indentity of tan:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( \alpha  +  \beta )

let α+β be t and transform:

\rm\displaystyle   \tan( \gamma ) =   -  \tan( t)

remember that tan(t)=tan(t±kπ) so

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm k\pi )

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm \pi )

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

\rm\displaystyle   \tan( \gamma ) =    \tan(   -\alpha  -\beta\pm \pi )

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm \pi

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ \pm \pi  }

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta \pm 0 )

likewise by Opposite Angle Identity we obtain:

\rm\displaystyle   \tan( \gamma ) =    \tan(   -\alpha   -\beta\pm 0 )

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\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm 0

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ 0  }

and we're done!

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