Respuesta :

frika

Answer:

(5,-4) and (-5,6)

Step-by-step explanation:

Given:

[tex]\left\{\begin{array}{l}x^2+x+y-26=0\\ \\x+y=1\end{array}\right.[/tex]

Solve it. First, express y in terms of x from the second equation:

[tex]y=1-x[/tex]

Substitute it into the first equation:

[tex]x^2+x+1-x-26=0\\ \\x^2-25=0\\ \\(x-5)(x+5)=0[/tex]

Apply zero product property:

[tex]x-5=0\ \text{or}\ x+5=0[/tex]

So,

[tex]x=5\ \text{or}\ x=-5[/tex]

When [tex]x=5,[/tex] then [tex]y=1-5=-4[/tex]

When [tex]x=-5,[/tex] then [tex]y=1-(-5)=6[/tex]

We get two solutions: (5,-4) and (-5,6)