Question 75804: The number 66 is divided into smaller numbers. One number is 3 more than twice the other number. Find the larger of the two numbers. Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! From the information in the problem we can write two equations. The first equation
comes from the fact that one number (call it x) and another number (call it y) add
up to be 66. So the first of our equations is:
.
x + y = 66
.
The second equation comes from the information that x is 3 more than twice y. In equation
form this can be written as:
.
x = 2y + 3
.
Substitute this into the first equation by replacing x by 2y + 3 and the first equation
becomes:
.
2y + 3 + y = 66
.
Subtract 3 from both sides for the purpose of eliminating the + 3 from the left side. When
you do this subtraction the equation becomes:
.
2y + y = 63
.
Add the terms on the left side and you get:
.
3y = 63
.
Divide both sides by 3 to find that:
.
y = 63/3 = 21
.
Since x + y = 66 we can substitute 21 for y to find that x + 21 = 66. Subtracting
21 from both sides we find that x = 66 - 21 = 45. So the larger of the two numbers
is 45.
.
Hope this helps you to understand that two equations can be used to solve this problem
and gives you some insight into the substitution method of solving the set of two equations.
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