Solve the system of equations 3r – 4s = 0 and 2r + 5s = 23. A. r = 12, s = 0 C. r = 4, s = 3 B. r = –12, s = 9 D. r = –1, s = –3
3r – 4s = 0, multiply 2 on both sides: 6r - 8s = 0 (equation 1) 2r + 5s = 23, multiply 3 on both sides: 6r + 15s = 69 (equation 2) equation 2 - equation 1 we get: 23s = 69, s= 69/23 = 3 3r - 4(3) = 0, r = 12/3 = 4 The solution for the system of equations 3r – 4s = 0 and 2r + 5s = 23 is C. r = 4, s = 3
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Asked 4/8/2014 5:24:08 AM
Updated 4/8/2014 8:04:28 AM
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3r – 4s = 0, multiply 2 on both sides: 6r - 8s = 0 (equation 1)
2r + 5s = 23, multiply 3 on both sides: 6r + 15s = 69 (equation 2)
equation 2 - equation 1 we get: 23s = 69, s= 69/23 = 3
3r - 4(3) = 0, r = 12/3 = 4
The solution for the system of equations 3r – 4s = 0 and 2r + 5s = 23 is C. r = 4, s = 3
Added 4/8/2014 8:04:28 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [4/8/2014 8:05:49 AM]

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Find the value of x in the equation 2(x – 3) + 5x = 5(2x + 6). A. 12 C. 2 B. –12 D. –2
Weegy: xUser: Find the possible values for s in the inequality 12s – 20 50 – 3s – 25. A. s 3 C. s 5/9 B. s 9 D. s 1/3 (More)
Updated 4/8/2014 6:54:09 AM
2(x – 3) + 5x = 5(2x + 6)
2x - 6 + 5x = 10x + 30
7x - 6 = 10x + 30
7x - 10x = 30 + 6
-3x = 36
x = -12
Added 4/8/2014 6:54:09 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [4/8/2014 7:06:32 AM]
Find the possible value or values of y in the quadratic equation 4 – 4y – y2 = 0. A. y = 2 C. y = –3, y = –1 B. y = –4, y = 1 D. y = –2 + 2 2 , y = –2 – 2 2
Weegy: 1 User: Find the possible value or values of r in the quadratic equation r2 – 7r – 8 = 0. (More)
Updated 4/8/2014 7:43:28 AM
r^2 – 7r – 8 = 0
(r + 1)(r - 8) = 0
r = -1 or 8
Added 4/8/2014 7:42:18 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [4/8/2014 7:43:13 AM]
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