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.
