The real numbers x and y satisfy the equations:
xy - x = 180 and y + xy = 208.
Let the two solutions be (x1, y1 and (x2, y2).
What is the value of x1 + 10y1 + x2 + 10y2

Answers

Answer 1

Answer:

The value of [tex]f = x_{1}+10\cdot y_{1}+x_{2}+10\cdot y_{2}[/tex] is 317.

Step-by-step explanation:

Let the system of nonlinear equations be:

[tex]x\cdot (y-1) = 180[/tex] (1)

[tex]y\cdot (1+x) = 208[/tex] (2)

From (1), we clear [tex]x[/tex]:

[tex]x = \frac{180}{y-1}[/tex]

Let apply this resulting expression in (2):

[tex]y\cdot \left(1+\frac{180}{y-1} \right)= 208[/tex]

[tex]y\cdot \left(\frac{y-1+180}{y-1} \right) = 208[/tex]

[tex]y\cdot \left(\frac{y+179}{y-1} \right)=208[/tex]  

[tex]y\cdot (y+179) = 208\cdot (y-1)[/tex]

[tex]y^{2}+179\cdot y = 208\cdot y-208[/tex]

[tex]y^{2}-29\cdot y+208 = 0[/tex]

By the Quadratic Formula we obtain the solution of the second order polynomial:

[tex]y_{1} = 16[/tex] and [tex]y_{2} = 13[/tex]

And the respective values of [tex]x[/tex] are, respectively:

[tex]y_{1} = 16[/tex]

[tex]x_{1} = \frac{180}{y_{1}-1}[/tex]

[tex]x_{1} = \frac{180}{15}[/tex]

[tex]x_{1} = 12[/tex]

[tex]y_{2} = 13[/tex]

[tex]x_{2} = \frac{180}{y_{2}-1}[/tex]

[tex]x_{2} = \frac{180}{12}[/tex]

[tex]x_{2} = 15[/tex]

If we know that [tex]x_{1} = 12[/tex], [tex]y_{1} = 16[/tex], [tex]x_{2} = 15[/tex] and [tex]y_{2} = 13[/tex], then the value of [tex]f = x_{1}+10\cdot y_{1}+x_{2}+10\cdot y_{2}[/tex] is:

[tex]f = 12+10\cdot (16)+15+10\cdot (13)[/tex]

[tex]f = 317[/tex]

The value of [tex]f = x_{1}+10\cdot y_{1}+x_{2}+10\cdot y_{2}[/tex] is 317.


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Answers

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Step-by-step explanation:

Number Rounded to Nearest Whole Number

4.645 5

5.645 6

6.645 7

7.645 8

8.645 9

9.645 10

10.645 11

11.645 12

12.645 13

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Answers

Answer:

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