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Sagot :
Let x be the larger number; Let y be the smaller number.
EQUATIONS:
x-2y=1
x+y+20=3x
---------
Rewrite as
x-2y=1
2x-y=20
==========
Multiply the 1st equation by 2 to get:
2x-4y=2
Subtract this result from the 1st equation to get:
3y=18
y=6
Substitute that into the 1st equation to get:
x-2*6=1
x=13
-------
Solution: x=13; y=6
EQUATIONS:
x-2y=1
x+y+20=3x
---------
Rewrite as
x-2y=1
2x-y=20
==========
Multiply the 1st equation by 2 to get:
2x-4y=2
Subtract this result from the 1st equation to get:
3y=18
y=6
Substitute that into the 1st equation to get:
x-2*6=1
x=13
-------
Solution: x=13; y=6
Answer:
13 and 6
Step-by-step explanation:
Let x = larger number
Let y = smaller number
The larger of two numbers is: x =
one more than twice the smaller: 2y + 1
x = 2y + 1 {equation 1}
The sum of the numbers is: x + y =
20 less than three times the larger: 3x - 20
x + y = 3x - 20
This can be simplified. Subtract 3x from both sides.
x - 3x + y = -20
-2x + y = -20 {equation 2}
Since we already know from equation 1 that x=2y+1, substitute that into equation 2
and solve for y:
-2(2y+1) + y = -20
-4y - 2 + y = -20
-3y - 2 = -20
-3y = -18
y = -18/(-3)
y = 6
x = 2y+1 = 2(6)+1 = 12 + 1 = 13
Your numbers are 13 and 6
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