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mezya [45]
4 years ago
12

In the system of equations below, which variable would it be easiest to solve for? x + 4 y = 14. 3 x + 2 y = 12 The easiest to s

olve for is x in the first equation. The easiest to solve for is y in the first equation. The easiest to solve for is x in the second equation. The easiest to solve for is y in the second equation.
Mathematics
1 answer:
Murljashka [212]4 years ago
5 0

Answer:

<h2>The easiest to solve for is x in the first equation</h2>

Step-by-step explanation:

Given the system of equation, x + 4 y = 14. and 3 x + 2 y = 12, to solve for x, we can use the elimination method of solving simultaneous equation. We need to get y first.

x + 4 y = 14............ 1 * 3

3 x + 2 y = 12 ............ 2  * 1

Lets eliminate x first. Multiply equation 1 by 3 and subtract from equation 2.

3x + 12 y = 42.

3 x + 2 y = 12

Taking the diffrence;

12-2y =42 - 12

10y = 30

y = 3

From equation 1, x = 14-4y

x = 14-4(3)

x = 14-12

x = 2

It can be seen that the easiest way to get the value of x is by using the first equation and we are able to do the substitute easily <u>because the variable x has no coefficient in equation 1 compare to equation 2 </u>as such it will be easier to make the substitute for x in the first equation.

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Ann's first option is a plot of land adjacent to a current park.
oksian1 [2.3K]

Given:

The equation for the area of the first option is:

x^2+200x=166400

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To find:

The side length of the current square park.

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We have,

x^2+200x=166400

It can be written as:

x^2+200x-166400=0

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First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
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