We will investigate the how to construct an algebraic equation and solve for a variable considering the word problem.
We will denote the cost of the cell phone at Store A as a variable:

Similarly, the cost of cell phone at store B is given as follows:

We will deconstruct the rest of the problem statement and translate into an algebraic equation that would relate the cost of cell phones at each store.
We are given that:

So if we translate the above statement in terms of ( x ) and the cost of cell phone at store B we have:

Therefore,

We will use the above equation and solve for the variable ( x ) by evaluating the right hand side of the " = " sign as follows:

Therefore, the cost of the cell phone at Store A is:
