The sum of three numbers is 21. One of the numbers is 2 more than twice
a second number and 2 less than the third number. Find the numbers.
x+y+z = 21
x = 2y+2
x = z-2
now just crank it out
1st. No. = X.
2nd No. = 2x + 2
3rd No. = (2x +2) + 2 = 2x + 4.
x + (2x+2) + (2x+4) = 21.
X = ?.
2x + 2 = ?
2x + 4 = ?.
To find the three numbers, let's assign variables to them:
Let's call the first number "x".
Let's call the second number "y".
And let's call the third number "z".
According to the given information:
1) The sum of the three numbers is 21, so we can write the equation:
x + y + z = 21
2) One of the numbers is 2 more than twice a second number, which can be written as:
x = 2y + 2
3) One of the numbers is 2 less than the third number, which can be written as:
x = z - 2
Now we have a system of three equations. We can solve this system of equations using substitution or elimination method.
Let's solve it using the substitution method:
1) Substitute the value of x from equation (2) into equation (1):
(2y + 2) + y + z = 21
3y + z = 19 ----(4)
2) Substitute the value of x from equation (3) into equation (1):
(z - 2) + y + z = 21
y + 2z - 2 = 21
y + 2z = 23 ----(5)
Now we have a system of equations (4) and (5).
To solve for y, let's multiply equation (4) by 2 and subtract equation (5) from it:
6y + 2z - 2y - 2z = 38 - 23
4y = 15
y = 15/4
Now that we have found the value of y as 15/4, we can substitute it back into one of the original equations (2) or (3) to find the other variables.
Using equation (2):
x = 2(15/4) + 2
x = 30/4 + 2
x = 32/4
x = 8
Using equation (3):
x = z - 2
8 = z - 2
z = 8 + 2
z = 10
So, the three numbers are:
x = 8
y = 15/4 (or 3.75 as a decimal)
z = 10