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2 years ago

4 - 4y = -2 - 9x - 4y = -3

Mathematics

1 answer:

arsen [322]2 years ago

7 0

Answer:

x = -1/3 , y = 3/2

Step-by-step explanation:

Solve the following system:

{4 - 4 y = -2

-9 x - 4 y = -3

Express the system in standard form:

{0 - 4 y = -6

-9 x - 4 y = -3

Express the system in matrix form:

(0 | -4

-9 | -4)(x

y) = (-6

-3)

Write the system in augmented matrix form and use Gaussian elimination:

(0 | -4 | -6

-9 | -4 | -3)

Swap row 1 with row 2:

(-9 | -4 | -3

0 | -4 | -6)

Multiply row 1 by -1:

(9 | 4 | 3

0 | -4 | -6)

Divide row 2 by -2:

(9 | 4 | 3

0 | 2 | 3)

Divide row 2 by 2:

(9 | 4 | 3

0 | 1 | 3/2)

Subtract 4 × (row 2) from row 1:

(9 | 0 | -3

0 | 1 | 3/2)

Divide row 1 by 9:

(1 | 0 | -1/3

0 | 1 | 3/2)

Collect results:

Answer: {x = -1/3 , y = 3/2

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How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$

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21 and 22 are complementary angles. The measure of Z1 is 45 The measure of Z2 is 3x°. Find the value of x. The figure is not dra

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Step-by-step explanation:

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