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3 years ago

Given the system of equations, what is the solution?5x - 4y = 7x = 5 - 3/2y

Mathematics

1 answer:

nignag [31]3 years ago

8 0

9514 1404 393

Answer:

(x, y) = (61/23, 36/23)

Step-by-step explanation:

We assume you intend the equations to be ...

5x -4y = 7
x = 5 -3/2y

Then in general form, they are ...

5x -4y -7 = 0

2x +3y -10 = 0

The solution using the "cross multiplication method" is ...

1/(15 +8) = x/(40+21) = y/(-14 +50)

(x, y) = (61/23, 36/23)

_____

<em>Comment on "cross multiplication method"</em>

When the coefficients from the general form equations are written as a 2×4 array with the first and last columns the same ...

$\left\{\begin{array}{cccc}5&-4&-7&5\\2&3&-10&2\end{array}\right\}$

the solution is 1/d1 = x/d2 = y/d3, where d1, d2, d3 are the determinants of successive pairs of columns.

d1 = (5)(3) -(2)(-4)
d2 = (-4)(-10) -(3)(-7)
d3 = (-7)(2) -(-10)(5)

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Solve (x − 2) = 3<br> Give your solutions correct to 3 significant figures.

Pachacha [2.7K]

Answer:

(x − 2) = 3

Remove the bracket

x - 2 = 3

Group the constants at one side of the equation

x = 2 + 3

x = 5

Therefore x = 5.00 to 3 significant figures.

Hope this helps

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3 years ago

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':

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

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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.

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