What is an x-value of a solution of the system of equations: x - y - 1 = 0 y = sqrt(x + 5)

Steps to Solve

1. Solve the first equation for $y$ We have the equation $x - y - 1 = 0$. Adding $y$ to both sides gives $x - 1 = y$. So, we can write $y = x - 1$.

2. Substitute the expression for $y$ into the second equation The second equation is $y = \sqrt{x+5}$. We substitute $y = x - 1$ into this equation, giving us $x - 1 = \sqrt{x+5}$.

3. Square both sides of the equation To eliminate the square root, we square both sides of the equation: $(x - 1)^2 = (\sqrt{x+5})^2$ $x^2 - 2x + 1 = x + 5$

4. Rearrange the equation into a quadratic equation Subtract $x$ and $5$ from both sides to set the equation to zero: $x^2 - 2x + 1 - x - 5 = 0$ $x^2 - 3x - 4 = 0$

5. Solve the quadratic equation We can factor the quadratic equation: $(x - 4)(x + 1) = 0$ So, $x = 4$ or $x = -1$.

6. Check the solutions in the original equations We need to check if these x-values satisfy the original equations since squaring can introduce extraneous solutions.

For $x = 4$: $y = x - 1 = 4 - 1 = 3$ $y = \sqrt{x+5} = \sqrt{4+5} = \sqrt{9} = 3$ So, $x = 4$ is a valid solution.

For $x = -1$: $y = x - 1 = -1 - 1 = -2$ $y = \sqrt{x+5} = \sqrt{-1+5} = \sqrt{4} = 2$ Since $y$ cannot be both $-2$ and $2$, $x = -1$ is not a valid solution.

7. Final Answer The only valid x-value is $x = 4$.