Liam wants to buy a car and pay for it in three installments. The total cost of the car is $29,000. Two times the first installment is $1,000 more than the sum of the third installment and three times the second installment. Liam must pay 15% interest on the second and the third installments: the interest will amount to $2,100. If x, y, and z represent the first, second, and third installments, respectively, identify the augmented matrices that model Liam's situation.
Answers
Given : The total cost of the car is $29,000 . Paid in three installments . x , y & z are the installments
To find : x, y, and z
Step-by-step explanation:
The total cost of the car is $29,000
Two times the first installment is $1,000 more than the sum of the third installment and three times the second installment
=> 2x = 1000 + z + 3y
=> z = 2x - 3y - 1000
x + y + z = 29000 + 2100
=> x + y + z = 31100
=> x + y + 2x - 3y - 1000 = 31100
=> 3x - 2y = 32100
1st installment at start paid = x
remaining = 29000 - x
(29000 - x) * 15 /100 = Interest for 1st
=> (29000 - x) 3/20
=> 4350 -0.15x
installment paid 2nd = y
{ (29000 - x) + 4350 - 0.15x - y }* 15/100 = Interest 2nd
{ (29000 - x) + 4350 - 0.15x - y } 3/20
= 5002.5 - 0.1725x - 0.15y
Interest of 1st installment + Interest of 2nd installment = 2100
=> 4350 -0.15x + 5002.5 - 0.1725x - 0.15y = 2100
=> 0.3225x + 0.15y = 7,252.5
=> 6.45x + 3y = 1,45,050
=> 12.9x + 6y = 290100
=> 4.3x + 2y = 96700
4.3x + 2y = 96700
3x - 2y = 32100
=> 7.3x = 128800
=> x ≈ 17644 ( 17,643.8356)
3x - 2y = 32100
=> y ≈ 10416 (10,415.7534 )
x + y + z = 31100
=> z ≈ 3040 ( 3,040.4110 )
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Given:
The total cost of the car is 29,000 dollars. Which is paid in three installments . x , y and z are the installments.
To Find:
The value of x, y and z.
Solving:
It is said that 2 times the 1 installment is $1,000 more than the sum of the 3 rd installment and 3 times the second installment.
= 2x = 1000 + z + 3y
Transposing:
= z = 2x - 3y - 1000
x + y + z = 29000 + 2100
= x + y + z = 31100
= x + y + 2x - 3y - 1000 = 31100
= 3x - 2y = 32100
First installment at start paid = x (We have assumed)
Remaining Installment which is left to pay = 29,000 - x
(29,000 - x) * 15 /100 = Interest for First
= (29000 - x) * 3/20
=> 4350 - 0.15x
Installment Paid Second = y
(29000 - x) + 4350 - 0.15x - y ) * 15/100 = Interest Second
(29000 - x) + 4350 - 0.15x - y ) * 3/20
= 5002.5 - 0.1725x - 0.15y
Interest of 1st installment + Interest of 2nd installment = 2100
Using this principle solving:
= 4,350 - 0.15x + 5,002.5 - 0.1725x - 0.15y = 2,100
= 0.3225x + 0.15y = 7,252.5
= 6.45x + 3y = 1,45,050
= 12.9x + 6y = 2,90,100
= 4.3x + 2y = 96,700
= 4.3x + 2y = 96700
= 3x - 2y = 32,100
= 7.3x = 1,28,800
= x ≈ 17644 ( 17,643.8356)
3x - 2y = 32,100
= y ≈ 10416 (10,415.7534 )
x + y + z = 31100
= z ≈ 3040 ( 3,040.4110 )