Twice the first number is eleven more than the sum of the other two numbers. The sum of twice the first number and three times the third number is one more than the second number. The second number is equal to the sum of the first number and the third number. Find the three numbers.
let the numbers be x, y, z
2x=y+z+11
2x-y-z=11
2x+3z=y+1
2x-y+3z=1
y=x+z
-x+y-z=0
2x-y-z=11-------------- 1
2x-y+3z =1--------------2
-x+y-z= 0-------------- 3
consider equation 1 &2
Multiply 1 by -1
Multiply 2 by 1
we get
-2x+y+z =-11
2x-y+3z =1
Add the two
4z=-10 -------------4
z==-2.5
consider equation 2 & 3
Multiply 2 by 1
Multiply 3 by 1
we get
2x-y+3z =1
-x+y+-z =0
Add the two
1x+2z= 1-------------5
plug value of z
x-5=1
x=6
plug value of x & z in 1
12-y+2.5=11
y=3.5
One-thirdthird of a number is 11 more than one-fourthfourth of the same number.
What is the number?
Let's represent the three numbers as variables: the first number is 'x', the second number is 'y', and the third number is 'z'.
From the given information, we can translate the problem into equations:
1) "Twice the first number is eleven more than the sum of the other two numbers." can be written as:
2x = y + z + 11
2) "The sum of twice the first number and three times the third number is one more than the second number." can be written as:
2x + 3z = y + 1
3) "The second number is equal to the sum of the first number and the third number." can be written as:
y = x + z
Now, we have a system of three equations. We can solve them simultaneously to find the values of 'x', 'y', and 'z'.
1) y = x + z (equation 3)
Substitute equation 3 into equations 1 and 2:
2x = (x + z) + z + 11
2x + 3z = (x + z) + 1
Simplify these equations:
x - 2z = 11 (equation 4)
-x + 2z = 1 (equation 5)
Add equations 4 and 5, which cancels out the 'x' terms:
0 = 12
There is no possible value of 'z' that satisfies this equation. Therefore, there is no solution to this problem.