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Mazyrski [523]
3 years ago
14

Solving systems of linear equations algbraically. ​

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

(3,-6)

Step-by-step explanation:

  [1]    x - 2y = 15

  [2]    2x + 4y = -18

Graphic Representation of the Equations :

   -2y + x = 15        4y + 2x = -18  

 

Solve by Substitution :

// Solve equation [1] for the variable  x

 [1]    x = 2y + 15

// Plug this in for variable  x  in equation [2]

  [2]    2•(2y+15) + 4y = -18

  [2]    8y = -48

// Solve equation [2] for the variable  y

  [2]    8y = - 48

  [2]    y = - 6

// By now we know this much :

   x = 2y+15

   y = -6

// Use the  y  value to solve for  x

   x = 2(-6)+15 = 3

Solution :

{x,y} = {3,-6}

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Answer:

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3 years ago
How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)
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Answer:

Make use of the fact that as long as \sin(\theta) \ne 0 and \cos(\theta) \ne 0:

\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.

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\sin^{2}(\theta) + \cos^{2}(\theta) = 1.

Step-by-step explanation:

Assume that \sin(\theta) \ne 0 and \cos(\theta) \ne 0.

Make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) and \csc(\theta) = (1) / (\sin(\theta)) to rewrite the given expression as a combination of \sin(\theta) and \cos(\theta).

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}.

Since \cos(\theta) \ne 0:

\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}.

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

By the Pythagorean identity, \sin^{2}(\theta) + \cos^{2}(\theta) = 1. Rearrange this identity to obtain:

\sin^{2}(\theta) = 1 - \cos^{2}(\theta).

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

Again, make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) to obtain the desired result:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}.

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