X = Smallest #.
4X = Largest @.
X+ 8X = 18.
9X = 18,
X = 2.
4X = 4*2 = 8.
14 - 10 = 4 = Middle #.
A. 1 unknown: X.
B. X = Smallest #.
4X = Largest #.
X + 8X = 18.
C.
The sum of 3 numbers is 14
the largest is 4 times the smallest
the sum of the smallest and twice the largest is 18
what are the 3 numbers? show work?
A. how many unknowns you have?
-how many equations need to be solve for that many unknown?
B. what are the equations?
C. used the system of equations using matrices.
4X = Largest @.
X+ 8X = 18.
9X = 18,
X = 2.
4X = 4*2 = 8.
14 - 10 = 4 = Middle #.
A. 1 unknown: X.
B. X = Smallest #.
4X = Largest #.
X + 8X = 18.
C.
B. To obtain the equations, we can use the information given in the problem statement.
1. The sum of the three numbers is 14.
This can be expressed as:
x + y + z = 14 ----- (Equation 1)
2. The largest number is 4 times the smallest.
This can be expressed as:
z = 4x ----- (Equation 2)
3. The sum of the smallest number and twice the largest number is 18.
This can be expressed as:
x + 2z = 18 ----- (Equation 3)
C. Now, let's solve the system of linear equations using matrices:
Step 1: Rearrange Equations 2 and 3 to eliminate 'z' variable.
Equation 2: z = 4x
Equation 3: x + 2z = 18
Multiply Equation 3 by (-2) to get -2x - 4z = -36.
Step 2: Rewrite the equations in matrix form.
The augmented matrix representing this system of equations is:
[ 1 1 1 | 14 ] ----- (Matrix A)
[ 4 0 -1 | 0 ] ----- (Matrix B)
[ 1 0 2 | 18 ] ----- (Matrix C)
Step 3: Perform row operations to find the solution.
- Multiply Row 1 by (-4) and add it to Row 2: Row 2 - 4 * Row 1 = [0 -4 -5 | -4]
- Multiply Row 1 by (-1) and add it to Row 3: Row 3 - Row 1 = [-1 -1 1 | 4]
- Multiply Row 2 by (-1/4): (1/4) * Row 2 = [0 1 5/4 | 1]
The new augmented matrix after these row operations is:
[ 1 1 1 | 14 ]
[ 0 -4 -5 | -4 ]
[ 0 -1 1 | 4 ]
Step 4: Continue row operations to simplify the matrix.
- Multiply Row 2 by (-1/4): (1/4) * Row 2 = [0 1 5/4 | 1]
- Multiply Row 2 by -1 and add it to Row 3: Row 3 - Row 2 = [0 0 1 | 3]
The new augmented matrix after these row operations is:
[ 1 1 1 | 14 ]
[ 0 1 5/4 | 1 ]
[ 0 0 1 | 3 ]
Step 5: Back-substitution to find the values of x, y, and z.
Equation 3: z = 3
Equation 2: y + (5/4)z = 1
Substitute z with 3 in Equation 2: y + (5/4) * 3 = 1
y + (15/4) = 1
y = 1 - (15/4)
y = -11/4
Substitute z = 3 and y = -11/4 in Equation 1: x + y + z = 14
x + (-11/4) + 3 = 14
x - 11/4 + 12/4 = 14
x + 1/4 = 14
x = 14 - 1/4
x = 55/4
So, the three numbers are:
x = 55/4
y = -11/4
z = 3
Therefore, the three numbers are 55/4, -11/4, and 3.