Mathematics, 27.01.2020 18:31, demetricejames

Find the values of x and y that satisfy both equations below.
$27 ^{x - 2} = 9^{y} \\ 32 ^{y - \frac{3}{5} } = \frac{1}{8 ^{2x} }$

Answers: 3

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Answers

answered: xmanavongrove55

The answer to your question is (4/3, -1)

Step-by-step explanation:

Convert both equations to the same base

$27^{x - 2} = 9^{y}$

27 = 3 x 3 x 3 = 3³

9 = 3 x 3 = 3²

Substitution

$3^{3(x - 2)} = 3^{2y}$

Simplify

3x - 6 = 2y Equation l

$32^{y - 3/5} = \frac{1}{8^{2x}}$

32 = 2 x 2 x 2 x 2 x 2 = 2⁵

8 = 2 x 2 x 2 = 2³

2 $^{5(y - 3/5)} = 2^{-3(2x)}$

Simplify

5y - 3 = -6x Equation ll

Equal to zero both equations

3x - 2y - 6 = 0 Equation l

6x + 5y - 3 = 0 Equation ll

Solve by elimination

Multiply equation l by 5 and equation ll by 2

15x - 10y - 30 = 0

12x + 10y - 6 = 0

27x -36 = 0

Solve for x

27x = 36

x = 36 / 27

Simplify x = 4/3

Find the y value

6(4/3) + 5y - 3 = 0

8 + 5y - 3 = 0

5y = -8 + 3

5y = -5

y = -5/5

y = -1

Solution

(4/3, -1)

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answered: Guest

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Find the values of x and y that satisfy both equations below.
[tex]27 ^{x - 2} = 9^{y} \\ 32 ^{...