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Aleks04 [339]
3 years ago
8

Can someone please help solve these simultaneous equations?

Mathematics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

a) x = 2, y = 1

b) x = 9, y = -9

Step-by-step explanation:

a)     2^{x+y} = 8

   ⇒ 2^{x+y} = 2^{3}

   ⇒ x+y = 3

and

      3^{2x-y} = 27

  ⇒ 3^{2x-y} = 3^{3\\

  ⇒ 2x-y = 3

By solving the above two equations, we get;

x = 2 and y = 1

b)     3^{y+x} = 9^{y+x}

   ⇒ 3^{y+x} = 3^{2y+2x}

   ⇒ x=-y

and

       2^{x-y} = 8^{x-3}

   ⇒ 2^{x-y} = 2^{3x-9}

   ⇒ 2x + y=9

By solving the above two equations, we get;

x = 9 and y = -9

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<h3>The worth after 4 years is $ 680.24</h3>

<em><u>Solution:</u></em>

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n = the number of times that interest is compounded per unit t

t = the time the money is invested

From given,

n = 1 ( since interest is compounded annually)

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t = 4

r = 8 \% = \frac{8}{100} = 0.08

<em><u>Substituting the values we get,</u></em>

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