# Solving Word Problems with the Help of Equations

With the help of equations we can solve many various problems from physics, economics and other sciences.

Let's describe the general scheme for solving such problems:

- Introduce new variables, i.e. denote unknown quantities which we need to find (or they are needed to find required quantitie) by letters x,y,z.
- Using introduced variables and conditions in the problem, set up system of equations (or one equation).
- Solve system of equations (or one equation) and from obtained solution choose only those solutions that make sense according to problem.
- If required quantities are not denoted by x,y,z, then using obtained solutions, find answer to the problem.

**Example 1**. Two workers, working together, can do some work in 6 hours. First worker, working alone, can do all work 5 hours than second worker, if second worker works alone. How many hours does each of them need to complete the work, working separately?

Before solving, note the following: productivity of labor, i.e. part of work (let's denote it by A), that is done in some time, and time (let's denote it by t) that is needed to complete all work are mutually inverse values, i.e. At=1.

Therefore, if we denote by x hours time, that is needed for the first worker to complete the work, then second worker needs (x+5) hours to complete the work. Also, part of work that first worker does in 1 hour equals `1/x`, for second worker this amount equals `1/(x+5)`.

According to the condition of the problem, they did all work in 6 hours. Part of work that first worker does in 6 hour equals `6/x`, for second worker this quantity equals `6/(x+5)`. Since in 6 hours they did all work, i.e. part of completed work equals 1, then we obtain equation `6/x+6/(x+5)=1`. Solution to this equation is x=-3 and x=10. But x=-3 is not applicable because time can't be negative. Thus, only x=10 is applicable.

This means that first worker needs 10 hours to complet the work, while second worker needs `10+5=15` hours.

**Example 2**. A vessel contains 54 liters of acid. Some amount of acid was poured out. Then vessel was filled up with water (amount of poured acid, equals amount of water that was filled up). Solution was mixed. Then, again, same amount of solution was poured out and then filled up with water. Now, vessel contains 24 liters of pure acis. How much acid was poured out first time?

Let x liters is amount of acid that was poured out first time. Then vessel contains `(54-x)` liters of acid. Filling up vessel with water, we obtained 54 liters of solution. Therefore, in one liter of solution there was `(54-x)/54` liters of acid (concentration of solution). Second time again x liters of solution were poured out. In this amount of solution there was `(54-x)/54*x` liters of acid.

Therefore, first time x liters of acis were poured out, second time `(54-x)/54*x` liters of acid were poured out. Togehter it equals `54-24=30` liters.

Thus, `x+(54-x)/54*x=30`. Solving it we obtain that `x_1=90,x_2=18`. Clearly, `x=90` is not applicable, because vessel can hold only 54 liters. Thus, 18 liters of acid were poured out first time.