Your equation should look like \( (x+c)=a \) or \( (x-c)=u \). 2) What are the solutions to the equation? Choose 1 answer: (A) \( x=5 \pm 4 \) (B) \( x=-5 \pm 4 \) (C) \( x=5 \pm 16 \) (D) \( x=-5 \pm 16 \)

To solve for \( x \) in the forms \( (x+c)=a \) or \( (x-c)=u \), isolate \( x \) by manipulating the equation. You'll typically subtract \( c \) or add it depending on the equation format. With a bit of creativity with numbers, the correct set of solutions springs forth from careful evaluation of \( \pm \) signs. If we examine the options, we see that in option (A) \( x=5 \pm 4 \) yields two potential solutions: \( x=9 \) or \( x=1 \), which aligns perfectly with the equation forms we’re discussing. So, that's our champ here! Now, let’s have some fun with real-world applications! These types of equations pop up in all sorts of practical scenarios, from calculating distances in physics to balancing chemical equations in chemistry. Understanding how to manipulate these expressions helps solve problems more efficiently, like determining flight routes or optimizing production processes in industries. The magic of algebra lies in its capacity to connect abstract numbers with tangible problems we face every day!