Question: The average of $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ is 10. Solve for $ y $. - RTA
Title: How to Solve for $ y $ When the Average of $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ Is 10
Title: How to Solve for $ y $ When the Average of $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ Is 10
When solving mathematical word problems in algebra, one common task is finding the value of a variable when the average of given expressions is known. In this article, we’ll break down how to find $ y $ when the average of $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ equals 10.
Understanding the Context
Understanding the Problem
We’re told that the average of three expressions — $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ — is 10. To find $ y $, we start by remembering that the average of three numbers is the sum divided by 3.
So the equation becomes:
$$
rac{(3y + 1) + (y + 7) + (5y - 2)}{3} = 10
$$
Image Gallery
Key Insights
Step 1: Combine Like Terms in the Numerator
Let’s simplify the expression inside the parentheses:
$$
(3y + 1) + (y + 7) + (5y - 2)
$$
Add the $ y $-terms:
$ 3y + y + 5y = 9y $
Add the constant terms:
$ 1 + 7 - 2 = 6 $
So the sum is $ 9y + 6 $. Now substitute back into the equation:
$$
rac{9y + 6}{3} = 10
$$
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Step 2: Simplify the Fraction
Divide numerator by 3:
$$
3y + 2 = 10
$$
Step 3: Solve for $ y $
Subtract 2 from both sides:
$$
3y = 8
$$
Divide both sides by 3:
$$
y = rac{8}{3}
$$
Final Answer
The value of $ y $ that makes the average of $ 3y + 1 $, $ y + 7 $, and $ 5y - 2 $ equal to 10 is
$$
oxed{ rac{8}{3}}
$$
