Can someone solve the system of equations

Mathematics

1 answer:

Sergio [31]3 years ago

6 0

Try this suggested solution:
1. if 'Equation2'-'Equation1', then
(5x+4y)-(5x-y)=42-27; ⇒ 5y=15; ⇒ y=3.
2. if to substitute y=3 into the Equation1, it is possible to find 'x':
5x-3=27; ⇒ 5x=30; ⇒ x=6.

answer: (6;3)

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$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

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Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]$

Step 2: Differentiate

Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'$
Derivative Rule [Basic Power Rule]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)$
Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$