Question 882983: when the lesser of two consecutive integers is added to 5 times the larger number, the result is 71. find the integers
Answer by Leaf W.(135)   (Show Source):
You can put this solution on YOUR website! Hello!
***
Since the integers are consecutive, the second will be exactly one greater than the first. Thus, you can call the first (the lesser) integer x, and the second (larger) one x + 1.
***
Let us model the statement "when the lesser . . . is added to 5 times the larger number, the result is 71" with an equation. Let us figure each part out piece by piece:
"the lesser of two consecutive integers": x
"the larger number": x + 1
"5 times the larger number": 5(x + 1)
"when the lesser of two consecutive integers is added to 5 times the larger number": x + 5(x + 1)
"when the lesser of two consecutive integers is added to 5 times the larger number, the result is 71": x + 5(x + 1) = 71
Therefore, our equation to model this statement is x + 5(x + 1) = 71
***
Now we simply solve the equation.
x + 5(x + 1) = 71
Distribute the 5 through the (x + 1): x + 5x + 5 =71
Add like terms: 6x + 5 = 71
Subtract 5 from both sides: 6x = 66
Divide both sides by 6: x = 11
***
Thus, x, which represents the value of the lesser integer, is equal to 11. Since the greater integer is x + 1, it is equal to 12.
***
Therefore, the integers are 11 and 12.
Let me know if you need any further clarification! =)
|
|
|
