1
UNIT EIGHT
UNIVERSITY OF THE PEOPLE
1201-01: COLLEGE ALGEBRA
Task 1
(i) Let the costs of the three distinct toys be represented by variables x, y, and z (in dollars). According to the
scenario, the total cost of the three toys is $25, which gives the equation:
x + y + z = 25. (1)
, 2
Furthermore, one of the toys (say the toy with cost z) costs $20 more than three times the combined price of the
other two toys. This relationship is expressed as:
z = 20 + 3(x + y). (2)
Equation (2) can be rearranged to standard linear form: -3x - 3y + z = 20. The third statement, that the total cost
multiplied by five equals $125, simply confirms that 5(x+y+z)=125, which is equivalent to equation (1) since
5×25=125. Therefore, the system of linear equations representing the situation is:
x + y + z = 25
-3x - 3y + z = 20
(ii) To determine the nature of the system, we examine the augmented matrix:
[ 1 1 1 | 25 ]
[ -3 -3 1 | 20 ]
Performing row operations: Replace row 2 with row 2 + 3 times row 1:
[ 1 1 1 | 25 ]
[ 0 0 4 | 95 ]
The second row gives 4z = 95, so z = 23.75. Substituting into the first row yields x + y = 1.25. The system has two
equations in three unknowns, and the coefficient matrix has rank 2, which is less than the number of unknowns.
Therefore, the system is consistent (since a solution exists) and dependent, meaning it has infinitely many
solutions.
(iii) From the reduced form, we have z = 23.75 and x + y = 1.25. Let x be a free parameter t. Then y = 1.25 - t, and
z = 23.75, where t can be any real number that yields non-negative costs (typically t ≥ 0 and 1.25 - t ≥ 0, so 0 ≤ t ≤
1.25). Thus the cost of the toy that satisfies the given condition is $23.75, while the other two toys together cost
$1.25, but their individual costs are not uniquely determined. For example, if t = 0.50, then the three costs are
$0.50, $0.75, and $23.75. Any pair of non-negative numbers summing to $1.25 is possible.
Task 2
We solve the following system using Gaussian elimination:
x-y+z=4
5x - y + z = 6
3x - y + 5z = 5
Write the augmented matrix:
[ 1 -1 1 | 4 ]
[ 5 -1 1 | 6 ]
[ 3 -1 5 | 5 ]
Step 1: Eliminate x from rows 2 and 3 using row 1.
Row2 → Row2 – 5×Row1: (5-5, -1-5(-1), 1-5×1, 6-5×4) = (0, 4, -4, -14)
UNIT EIGHT
UNIVERSITY OF THE PEOPLE
1201-01: COLLEGE ALGEBRA
Task 1
(i) Let the costs of the three distinct toys be represented by variables x, y, and z (in dollars). According to the
scenario, the total cost of the three toys is $25, which gives the equation:
x + y + z = 25. (1)
, 2
Furthermore, one of the toys (say the toy with cost z) costs $20 more than three times the combined price of the
other two toys. This relationship is expressed as:
z = 20 + 3(x + y). (2)
Equation (2) can be rearranged to standard linear form: -3x - 3y + z = 20. The third statement, that the total cost
multiplied by five equals $125, simply confirms that 5(x+y+z)=125, which is equivalent to equation (1) since
5×25=125. Therefore, the system of linear equations representing the situation is:
x + y + z = 25
-3x - 3y + z = 20
(ii) To determine the nature of the system, we examine the augmented matrix:
[ 1 1 1 | 25 ]
[ -3 -3 1 | 20 ]
Performing row operations: Replace row 2 with row 2 + 3 times row 1:
[ 1 1 1 | 25 ]
[ 0 0 4 | 95 ]
The second row gives 4z = 95, so z = 23.75. Substituting into the first row yields x + y = 1.25. The system has two
equations in three unknowns, and the coefficient matrix has rank 2, which is less than the number of unknowns.
Therefore, the system is consistent (since a solution exists) and dependent, meaning it has infinitely many
solutions.
(iii) From the reduced form, we have z = 23.75 and x + y = 1.25. Let x be a free parameter t. Then y = 1.25 - t, and
z = 23.75, where t can be any real number that yields non-negative costs (typically t ≥ 0 and 1.25 - t ≥ 0, so 0 ≤ t ≤
1.25). Thus the cost of the toy that satisfies the given condition is $23.75, while the other two toys together cost
$1.25, but their individual costs are not uniquely determined. For example, if t = 0.50, then the three costs are
$0.50, $0.75, and $23.75. Any pair of non-negative numbers summing to $1.25 is possible.
Task 2
We solve the following system using Gaussian elimination:
x-y+z=4
5x - y + z = 6
3x - y + 5z = 5
Write the augmented matrix:
[ 1 -1 1 | 4 ]
[ 5 -1 1 | 6 ]
[ 3 -1 5 | 5 ]
Step 1: Eliminate x from rows 2 and 3 using row 1.
Row2 → Row2 – 5×Row1: (5-5, -1-5(-1), 1-5×1, 6-5×4) = (0, 4, -4, -14)
