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3 years ago

Find the value of x and y, (5x-3y, 3x-y) = (16, 12)

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

x = 10 / 7 (or) 1.43 <rounded to 2 decimal place>

y = - 54 / 7 (or) -7.71 <rounded to 2 decimal place>

Step-by-step explanation:

Another way to rewrite (5x-3y, 3x-y) = (16, 12),

5x - 3y = 16 -- (1)

3x - y = 12 -- (2)

Rewriting (1), y = -5/3x - 16/3 -- (1)'

Substituting (1)' into (2),

3x + 5/3x + 16/3 = 12

14/3x = 20/3

x = 10 / 7 -- (3)

Substituting (3) into (2),

3(10/7) - y = 12

30/7 - y = 12

-y = 54 / 7

y = - 54/7-- (4)

Therefore,

x = 10 / 7 (or) 1.43 <rounded to 2 decimal place>

y = - 54 / 7 (or) -7.71 <rounded to 2 decimal place>

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
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Subtraction

Left to Right

Algebra I

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Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
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Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
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Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$