Respuesta :

From the secon equation, derive


[tex] y = 10-x [/tex]


Substitute in the first equation:

[tex]x^2+(10-x)^2 = 58[/tex]
Expand and simplify:
[tex]x^2 - 10 x + 21 = 0[/tex]
Solve with the quadratic equation to get [tex]x=3[/tex] or [tex]x=7[/tex]
Derive back the values for [tex] y [/tex]:
[tex]x=3 \implies y=7[/tex]
[tex]x=7 \implies y=3[/tex]
which makes sense, since the problem is symmetric in [tex] x [/tex] and [tex] y [/tex]












DeanR

That's the intersection of a circle and a line.


[tex]y = 10 - x[/tex]


[tex]x^2 + (10 - x)^2 = 58[/tex]


[tex]x^2 + 100 -20 x + x^2 = 58[/tex]


[tex]2x^2 -20 x + 42 = 0[/tex]


[tex]x^2 -10 x + 21 = 0[/tex]


[tex](x-7)(x-3)=0[/tex]


[tex]x = 7 \quad \textrm{or} \quad x=3[/tex]


By the symmetry of the problem when x is one solution y is the other:


(x,y) = (7,3) or (3,7)


Check:


[tex]7^2+3^2=49+9=58 \quad\checkmark[/tex]


Otras preguntas