Find the two numbers whose arithmetic mean is 49, where one number is 17 more than twice the other.
We recall that in order to calculate the arithmetic mean of a data set, we divide the sum of the values by the number of values. If we let our two numbers in this case be 𝑥 and 𝑦, then their arithmetic mean is equal to 𝑥 plus 𝑦 divided by two. We are told this is equal to 49. Multiplying both sides of this equation by two gives us 𝑥 plus 𝑦 is equal to 98. The sum of our two numbers is 98. We will call this equation one.
We are also told that one number is 17 more than twice the other. If we assume that 𝑦 is the larger number, then 𝑦 is equal to two 𝑥 plus 17. We will call this equation two, and we now have a set of simultaneous equations that we can solve by substitution or elimination. As 𝑦 is already the subject of equation two, it makes sense to substitute this into equation one. Replacing 𝑦 with two 𝑥 plus 17 gives us 𝑥 plus two 𝑥 plus 17 is equal to 98.
The left-hand side simplifies to three 𝑥 plus 17. We can then subtract 17 from both sides of the new equation. Three 𝑥 is equal to 81. Dividing both sides of this equation by three gives us a value of 𝑥 of 27. We can now substitute this value back into equation one or equation two. Substituting 𝑥 equals 27 into equation two gives us 𝑦 is equal to two multiplied by 27 plus 17. Double 27 is 54. 𝑦 is, therefore, equal to 54 plus 17. This gives us an answer of 71. We could check this in equation one as the sum of 27 and 71 is 98.
The two numbers whose arithmetic mean is 49, where one number is 17 more than twice the other, are 27 and 71.