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Goshia [24]
3 years ago
12

Solve the system of equations. 3x+5y=9 3x+2y=3

Mathematics
1 answer:
KIM [24]3 years ago
5 0
3x+5y=9           3x+5y=9
-1(3x+2y=3)     -3x-2y=-3
                         3y=6
                          y=2
3x+5(2)=9
3x+10=9
     -10   -10
3x=-1
x=-1/3 (-1/3,2)

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3 years ago
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4 Tan A/1-Tan^4=Tan2A + Sin2A​
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… = 1/cos(2<em>A</em>) (sin(2<em>A</em>) + sin(2<em>A</em>) cos(2<em>A</em>))

• expand the functions of 2<em>A</em> using the double angle identities

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• factor out sin(<em>A</em>) cos(<em>A</em>)

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• simplify the last factor using the Pythagorean identity, 1 - sin²(<em>A</em>) = cos²(<em>A</em>)

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• rearrange terms in the product

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• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(<em>A</em>)

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… = 4 sin(<em>A</em>) cos(<em>A</em>) / (2 - sec²(<em>A</em>))

• divide through again by cos²(<em>A</em>)

… = (4 sin(<em>A</em>)/cos(<em>A</em>)) / (2/cos²(<em>A</em>) - sec²(<em>A</em>)/cos²(<em>A</em>))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(<em>A</em>) / (2 sec²(<em>A</em>) - sec⁴(<em>A</em>))

• factor out sec²(<em>A</em>) in the denominator

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(Note that some of these steps are optional or can be done simultaneously)

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Answer:

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