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## Solved Examples

**Question 1:**Find the numbers such that one-third of the first number added to twice of the second number gives 2 and four times of first number added to the five times the second number gives 5.

**Solution:**

Let the two numbers be x and y

Step 1:

Given,

One-third of the first number + twice of the second number = 2

=> $\frac{1}{3}$x + 2y = 2

=> x + 6y = 6 ...................................(1)

Also,

Four times of first number + five times the second number = 5

=> 4x + 5y = 5 .......................................(2)

Step 2:

(1) => x = 6 - 6y, put in equation (2)

=> 4(6 - 6y) + 5y = 5

=> 24 - 24y + 5y = 5

=> - 19y = 5 - 24 = - 19

=> y = 1

Step 3:

Put y = 1 in x = 6 - 6y

=> x = 6 - 6 * 1 = 6 - 6 = 0

=> x = 0

The solution to the system is

Step 1:

Given,

One-third of the first number + twice of the second number = 2

=> $\frac{1}{3}$x + 2y = 2

=> x + 6y = 6 ...................................(1)

Also,

Four times of first number + five times the second number = 5

=> 4x + 5y = 5 .......................................(2)

Step 2:

(1) => x = 6 - 6y, put in equation (2)

=> 4(6 - 6y) + 5y = 5

=> 24 - 24y + 5y = 5

=> - 19y = 5 - 24 = - 19

=> y = 1

Step 3:

Put y = 1 in x = 6 - 6y

=> x = 6 - 6 * 1 = 6 - 6 = 0

=> x = 0

The solution to the system is

**x = 0 and y = 1**.**Question 2:**Amren starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was 1500 dollars after 4 years of service and 1800 dollars after 10 years of service, what was his starting salary and what is the annual increment.

**Solution:**

Let Amren's staring salary = x dollar

Annual increment = y dollars

=> x + 4y = 1500 .....................(1)

and

=> x + 10y = 1800 .......................(2)

Subtracting equation (1) from equation (2)

=> x + 10y - ( x + 4y) = 1800 - 1500

=> x + 10y - x - 4y = 300

=> 6y = 300

=> y = $\frac{300}{6}$

=> y = 50

Put y = 50 in equation (1)

=> x + 4 * 50 = 1500

=> x + 200 = 1500

=> x = 1500 - 200 = 1300

=> x = 1300

Annual increment = y dollars

Step 1:Step 1:

**Salary in 4 years:**

=> x + 4y = 1500 .....................(1)

and

**salary in 10 years:**

=> x + 10y = 1800 .......................(2)

**Step 2:**

Solve for x and y,Subtracting equation (1) from equation (2)

=> x + 10y - ( x + 4y) = 1800 - 1500

=> x + 10y - x - 4y = 300

=> 6y = 300

=> y = $\frac{300}{6}$

=> y = 50

Put y = 50 in equation (1)

=> x + 4 * 50 = 1500

=> x + 200 = 1500

=> x = 1500 - 200 = 1300

=> x = 1300

**Answer:**

**Starting salary of Amren = 1300 dollars and annual increment = 50 dollars.**