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2 years ago

Which method would be best (quickest) for solving the system below: 3x - 4y = -2 y = 2x + 1

1 answer:

8 0

To solve with Elimination:

Write the equations under one another, like this:

2x - y = -1
+ 3x + 4y = 26

Ideally, we would like for one of the variables to be eliminated when we add vertically (straight down). But if we add them as they are this does not happen. We must manipulate one of the equations so that it will happen. Again, you can try to eliminate either x or y. I always look for a term that has a coefficient of 1 (or negative 1). So, let's use that y from the first equation again.

If the coefficient of the y in the other equation is POSITIVE 4, then I need the coefficient from the first equation to be its opposite, NEGATIVE 4. To do this, simply multiply the first equation by 4, this will create MAGIC!

4( 2x - y = -1)
+ 3x + 4y = 26

Be certain to Distribute across the entire first equation, so multiply all three terms by 4.

8x - 4y = -4
+ 3x + 4y = 26

Now add straight down (vertically). The y term will be eliminated.

11x = 22

Divide both sides of the equation by 11.

x = 2

Almost there! Now, substitute the 2 in for x in either of the original equations. Either one will work. I'm gonna use the second equation.

3x + 4y = 26

3(2) + 4y = 26

6 + 4y = 26

Subtract 6 from both sides of the equation.

4y = 20

Divide both sides of the equation by 4.

y = 5

That's it! There it is again. Put it all together. If x = 2 and y = 5, then the solution is the ordered pair, (2,5).

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Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

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Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

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