The given system of equation is,
[tex]\begin{gathered} y=x^2+5x-6\text{ ------(1)} \\ x+y=10\text{ ------(2)} \end{gathered}[/tex]
Rewrite equation (2).
[tex]y=10-x[/tex]
Now, put y=10-x in equation (1).
[tex]10-x=x^2+5x-6[/tex]
Rewrite the above equation and solve for x.
[tex]\begin{gathered} 0=x^2+5x-6-10+x \\ 0=x^2+6x-16 \\ 0=x^2+8x-2x-8\times2 \\ 0=x(x+8)-2(x+8) \\ 0=(x-2)(x+8) \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} x-2=0\text{ or x+8=0} \\ x=2\text{ or x=-8} \end{gathered}[/tex]
x=-8 is not in the first quadrant.
Put x=2 in equation (2) to find the corresponding value of y.
[tex]\begin{gathered} y=10-2 \\ y=8 \end{gathered}[/tex]
The coordinate (x, y)=(2, 8) is in the first quadrant.
So, the x coordinate of the solution located in the first quadrant is x=2 .
