Question

The following system has to real solutions. What is the x-coordinate of the solution located in the 1st quadrant?if necessary, round your answer to the nearest integer.

Answer 1

The given system of equation is,

`\begin{gathered} y=x^2+5x-6\text{ ------(1)} \\ x+y=10\text{ ------(2)} \end{gathered}`

Rewrite equation (2).

`y=10-x`

Now, put y=10-x in equation (1).

`10-x=x^2+5x-6`

Rewrite the above equation and solve for x.

`\begin{gathered} 0=x^2+5x-6-10+x \\ 0=x^2+6x-16 \\ 0=x^2+8x-2x-8*2 \\ 0=x(x+8)-2(x+8) \\ 0=(x-2)(x+8) \end{gathered}`

Hence,

`\begin{gathered} x-2=0\text{ or x+8=0} \\ x=2\text{ or x=-8} \end{gathered}`

x=-8 is not in the first quadrant.

Put x=2 in equation (2) to find the corresponding value of y.

`\begin{gathered} y=10-2 \\ y=8 \end{gathered}`

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 .