SOLUTION: The sum of two numbers is 73. The second is 7 more than 5 times the first. What are the two numbers?
Can some please help out with this question.
v/r
Jess
Algebra.Com
Question 148180: The sum of two numbers is 73. The second is 7 more than 5 times the first. What are the two numbers?
Can some please help out with this question.
v/r
Jess
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The sum of two numbers is 73. The second is 7 more than 5 times the first. What are the two numbers?
Let's call them N1 and N2.
N1 + N2 = 73
N2 = 5*N1 + 7
----------
Sub for N2 in the 1st equation
N1 + 5*N1+7 = 73
6*N1 + 7 = 73
6*N1 = 66
N1 = 11
-----
N2 = 73-11 = 62
-------
It all fits.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of two numbers is 73. The second is 7 more than 5 times the first. What are the two numbers?
----------------------
Using two variables:
EQUATIONS:
x + y = 73
y = 5x + 7
-------------
Substitute the 2nd into the 1st to get:
x + (5x+7) = 73
6x = 66
x = 11
-------
Substitute into y = 5x+7 to solve for "y":
y = 5*11+7
y = 62
===================
Using one variable:
Let 1st be "x";
then 2nd = 5x+7
EQUATION:
x + 5x+7 = 73
6x + 7 = 73
6x = 66
x = 11
then y = 62
=============
Cheers,
Stan H.
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