12a - 5b = 18
15a - 5b = 30
Subtract the first equation from the second:
3a = 12
Divide by 3
a = 4
Substitute into either equation and solve for b.
12(4) - 5b = 18
48 - 5b = 18
-5b = -30
b = 6
The solution to the system is a = 4 and b = 6.
2007-07-22 08:15:23
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answer #1
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answered by suesysgoddess 6
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If you do the second equation minus the first equation:
15a - 12a = 3a
-5b - (-5b) = 0
30 - 18 = 12
So: 3a = 12 => a = 12/3 = 4
12*4 - 5b = 18
48 - 5b = 18
48 - 18 = 30 = 5b
b = 30/5 = 6
2007-07-22 08:10:19
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answer #2
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answered by claudeaf 3
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12a - 5b = 18
- 15a + 5b = - 30----ADD
-3a = - 12
a = 4
- 60 + 5b = - 30
5b = 30
b = 6
Solution is a = 4 , b = 6
2007-07-22 08:13:00
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answer #3
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answered by Como 7
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1st equation:
12a - 5b = 18
12a = 18 + 5b
2nd equation:
15(18 + 5b) - 5b = 30
270 + 75b - 5b = 30
70b = - 240
b = - 7/24
1st equation (substitute b):
12a = 18 + 5(- 7/24)
12a = 18 - 35/24
12a = 397/24
a = 397/288 or 1 109/288
Answer: a = 397/288 or 1 109/288, b = - 7/24
Proof:
12(397/288) - 5(- 7/24) = 18
397/24 + 35/24 = 18
2007-07-26 00:30:10
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answer #4
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answered by Jun Agruda 7
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a million. purely upload the two equations to a minimum of one yet another: 4x + 0y = 40 x = 10, and consequently y = -2 2. Subtract the best equation from the backside equation 3a - 0b = 12 a = 4, and consequently b = 6 3. Substitution: n = 2m + 3 from first eq, and substituting into 2nd: 3m = 2(2m + 3) - 9 3m = 4m - 3 m = 3, and consequently n = 2*3 + 3 = 9
2016-11-10 02:57:45
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answer #5
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answered by ? 4
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12a-5b=18
- 15a-5b=30
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-3a=-12
a=4*
12*4-5b=18
48-5b=18
-5b=-30
b=6*
2007-07-22 08:48:14
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answer #6
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answered by CoolioMADDog 4
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