## College Algebra (6th Edition)

$x=4$ satisfies the given conditions
$y=x+\sqrt{x+5}$ and $y=7$ Substitute $y$ by $7$: $x+\sqrt{x+5}=7$ Take $x$ to the right side: $\sqrt{x+5}=7-x$ Square both sides of the equation: $(\sqrt{x+5})^{2}=(7-x)^{2}$ $x+5=49-14x+x^{2}$ Take all terms to the right side: $x^{2}-14x+49-5-x=0$ $x^{2}-15x+44=0$ Solve by factoring: $(x-11)(x-4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-11=0$ $x=11$ $x-4=0$ $x=4$ Check the solutions found by plugging them into the original equation: $x=11$ $11+\sqrt{11+5}=7$ False $x=4$ $4+\sqrt{4+5}=7$ True $x=4$ satisfies the given conditions