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Hoochie [10]
3 years ago
10

4x-2y = 2 3x-2y=-19 Help

Mathematics
2 answers:
Anon25 [30]3 years ago
6 0

Answer:

{x,y}={21,41}

Step-by-step explanation:

Step by Step Solution:

More Icon

System of Linear Equations entered :

  [1]    4x - 2y = 2

  [2]    3x - 2y = -19

Graphic Representation of the Equations :

   -2y + 4x = 2        -2y + 3x = -19  

 

Solve by Substitution :

// Solve equation [2] for the variable  x

 [2]    3x = 2y - 19

 [2]    x = 2y/3 - 19/3

// Plug this in for variable  x  in equation [1]

  [1]    4•(2y/3-19/3) - 2y = 2

  [1]    2y/3 = 82/3

  [1]    2y = 82

// Solve equation [1] for the variable  y

  [1]    2y = 82

  [1]    y = 41

// By now we know this much :

   x = 2y/3-19/3

   y = 41

// Use the  y  value to solve for  x

   x = (2/3)(41)-19/3 = 21

Solution :

{x,y} = {21,41}

Dominik [7]3 years ago
4 0

Answer: {x,y}={21,41}

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Let

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and subtracting this from S_n gives

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\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

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\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

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\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

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