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)