SOLUTION: To Solve Integer problems:
The sum of two integers is ten. Three times the larger integer is three less than 8 times the small integer. Find the integers.
my try:
Firs
Algebra.Com
Question 192253This question is from textbook Intermediate Algeba
: To Solve Integer problems:
The sum of two integers is ten. Three times the larger integer is three less than 8 times the small integer. Find the integers.
my try:
First integer: 3n
Second integer: 8n-3
3n + 8n-3 = 10
11n -3 =10
11n = 13
n = 13/11
Not right?
Help!
This question is from textbook Intermediate Algeba
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The sum of two integers is ten. Three times the larger integer is three less than 8 times the small integer. Find the integers.
my try:
First integer: 3n
Second integer: 8n-3
3n + 8n-3 = 10
11n -3 =10
11n = 13
n = 13/11
Not right?
Help!
------------
Not much chance it's right since it's not an integer.
smaller integer = s
bigger integer = b
3b+3 = 8s
b+s = 10 --> s = 10-b
Sub for s into 1st eqn
3b+3 = 8*(10-b)
3b+3 = 80-8b
11b = 77
b = 7
s = 3
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of two integers is ten. Three times the larger integer is three less than 8 times the small integer. Find the integers.
---
Equations:
x + y = 10
3x = 8y-3
---------------
Substitute to solve for "x":
3x = 8(10-x)-3
3x = 80 - 8x -3
11x = 77
x = 7 (one of the integers)
-----------
Substitute to solve for "y":
7 + y = 10
y = 3(the other integer)
=============================
Cheers,
Stan H.
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