This is the third part of the Solving Number Problems Series. The first part can be read here and the second part can be read here. In this post I will continue worked examples using problems which are slightly more complicated than the problems in the previous two parts. Without further ado, lets start with the seventh problem in the series.
Twice a number added to is times that number. What is the number?
In the two previous posts, we have learned that if is a number, then twice that number is . So, twice a number added to is represented by . Now, that number, the is five times that number of . So, we can now set up the equation
If we solve for , we have .
Checking the Answer
Twice added to is . Five times is . So, we are correct.
Four times the sum of a number and is . What is the number?
This one is quite tricky. Some interpret this as . But that is not correct. If you read it carefully, the equation is “four times a number added to three.” However, what we want is four times the sum of a number and 3. So, if we let be the number, then the sum of a number and 3 is . Four times the sum of a number and 3 is there . Now, we set up the equation
Solving, we have
This gives us and .
Checking the Answer
Four times the sum of and is four times which is equal to . So, we are correct.
The sum of three numbers is . The first number is less than the second number. The third number is four times the second number.
As you can observe, the second number has no description, so we let it be . The first number is less than the second number by , so the first number is . The third number is four times the second number or .
Their sum is , so
Simplifying, we have
So and the first number is . The third number is .
Checking the Ansewer
Left as an excersise.
You have probably noticed that in solving word problems, it is important to accurately convert phrases into algebraic expressions/equations and vice versa. The word “is” for instance is the same as “equal.” Of course, these keywords are help us in understanding the problem, but we should remember to understand the problem as a whole. In the fourth part (last part) of this series, we will discuss more complicated problems.