I have been on this for at least 3 hours I cant figure it out!! I am going insane!!! someone please help me:
1.) solve this system of equations
4x+7y2z=0
3x5y+3z=9
3x+6yz=1
2.)Daniel made an initial deposit of $85 into a savings account that earns $2 interest each week. Bridget made a deposit of $90 into an account that earns $1 interest each week. If neither Daniel or Bridget make another deposit, after how many weeks will their accounts be equal?
3.)2x – 5y = –12
3x – 4y = 3
Solve the sys. of equations above using the elimination method above
4.) Solve using substitution method:
5x – y = –2
10x – 3y = –7
please i would really appreciate your help!!! i have been on this for hours!!! i just cant get it!
2 Answers

1. multiply the 3rd by 3 and add to the 2nd–> new 2nd , multiply the 3rd by 2 and subtract from 1st –> new 1st , multiply new 1st by 6 and add to new 2nd , find y , then x , then z
#2. 5 weeks.. 90 85 = 5 and 2 1 = 1 { net gain to $85 }…5/1 = 5
#3. multiply the 1st by 3/2 and subtract from the 2nd , solve for y , then x
#4. solve the 1st for y , put into the 2nd ..solve for x , then y…you do the work

1)
4x +7y – 2z = 0 — (1)
3x – 5y + 3z = 9 — (2)
3x + 6y – z = 1 — (3)
From (3), we have: z = 3x + 6y – 1 — (4)
Substituting (4) into (1), we have:
4x + 7y – 2(3x + 6y – 1) = 0
4x + 7y – 6x – 12y + 2 = 0
2x – 5y + 2 = 0
2x + 5y – 2 = 0 — (5)
Substituting (4) into (2), we have:
3x – 5y + 3(3x + 6y – 1) = 9
3x – 5y + 9x + 18y – 3 = 9
12x + 13y = 12
13y = 12 – 12x
y = 12/13 – (12/13)x — (6)
Substituting (6) into (5), we have:
2x + 5(12/13 – (12/13)x) – 2 = 0
26x + 5(12 – 12x) – 26 = 0
26x + 60 – 60x – 26 = 0
34x = 34
x = 1
Substituting x = 1 into (6), we have:
y = 12/13 – (12/13)x
y = 12/13 – 12/13
y = 0
Substituting x = 1 and y = 0 into (4), we have:
z = 3(1) + 6(0) – 1
z = 2
Hence, we have x = 1, y = 0 and z = 2
2)
Let the number of weeks be x
85 + 2x = 90 + x
2x – x = 90 – 85
x = 5
After 5 weeks their accounts will be equal.
3)
2x – 5y = 12 — (1)
3x – 4y = 3 — (2)
From (1), we have:
4(2x – 5y) = 4(12)
8x – 20y = 48 — (3)
From (2), we have:
5(3x – 4y) = 5(3)
15x – 20y = 15 — (4)
Subtracting (3) from (4), we have:
(15x – 20y) – (8x – 20y) = 15 – (48)
7x = 63
x = 9
From (1), we have:
2(9) – 5y = 12
18 – 5y = 12
5y = 30
y = 6
4)
5x – y = 2 — (1)
10x – 3y = 7 — (2)
From (1), we have:
y = 5x + 2 — (3)
Substituting (3) into (2), we have:
10x – 3(5x + 2) = 7
10x – 15x – 6 = 7
5x = 1
x = 1/5
Substituting x = 1/5 into (3), we have:
y = 5(1/5) + 2
y = 1 + 2
y = 3