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3 years ago

Find algebraically in RxR using substituting method the S.S. of the two equations:

Mathematics

2 answers:

Art [367]3 years ago

8 0

Answer:

x=3 y=1

Step-by-step explanation:

x+y=4

y=-x+4

plug in the y

2x-(-x+4)=5

2x+x-4=5

3x=9

x=3

plug in x=3 into first equation to find y

y=-(3)+4

y=-3+4

y=1

nadya68 [22]3 years ago

6 0

Answer:

Step-by-step explanation:

x + y = 4 equation 1

2x - y = 5 equation 2

using substitution method,

x= 4 - y equation 3

substitute equation 3 into equation 2

2(4-y) -y = 5

8 - 2y - y = 5

8 - 3y = 5

-3y = 5 - 8

-3y = -3

divide through by -3

-3y/-3 = -3/-3

y = 1

substitue the value of y into equation 1

x + y = 4

x + 1 = 4

x = 4 - 1

x = 3

Hence, x = 3 and y = 1

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The range is the output of the function, and there are many ways to find it and write it. Let's find it first by plugging in all the domain values (domain means input) into the function:
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or, subtract the smallest value from the largest one:
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So, there are those ways to write it, there are more, but I think you should stick with the first way because that's how the problem was presented to you. If you have any questions, hmu!

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Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

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Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

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$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

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