3a + 5c=38
7a + 2c=50
2007-03-10 14:44:16 · 4 answers · asked by Michele N 1 in Science & Mathematics ➔ Mathematics
Multiyplying the first equation by 2, and the second by 5:
6a + 10c = 76
35a + 10c = 250
Subtracting the above equations:
29a = 174
a = 6
Substituting this value in (3a+5c=38)
18+5c = 38
5c=20
c=4
Hence a=6, c=4
2007-03-10 14:50:20 · answer #1 · answered by Shrey G 3 · 1⤊ 0⤋
3a + 5c = 38
7a + 2c = 50
Let's using elimination to solve this. Multiply the top equation by 7 and the bottom equation by 3.
21a + 35c = 266
21a + 6c = 150
Subtract equations,
29c = 116
Solve for c.
c = 116/29 = 4
Now that we have c = 4, we can easily get the other value by plugging this into one of the equation.
3a + 5c = 38. c = 4, so
3a + 5(4) = 38
3a + 20 = 38
3a = 18
a = 6
a = 6, c = 4
2007-03-10 22:49:08 · answer #2 · answered by Puggy 7 · 0⤊ 0⤋
I) (3a + 5c=38 (-2)
II) (7a + 2c=50 (5)
I) (-6a - 10c= -76
II) (35a + 10c= 250 (5)
29a = 174
a = 174/29
a = 6
3a + 5c=38
3(6) + 5c = 38
5c = 38 - 18
5c = 20
c = 20 : 5
c = 4
Answer: a = 6 and c = 4
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2007-03-10 23:01:20 · answer #3 · answered by aeiou 7 · 0⤊ 0⤋
(3a + 5c = 38) x 2
(7a + 2c = 50) x 5
6a + 10c = 76
35a + 10c = 250
29a = 174
a = 6
c = (50 - 7a)/2
c = (50 - 7(6))/2
c = (50 - 42)/2 = 8/2 = 4
a = 6 and c = 4
2007-03-10 22:50:46 · answer #4 · answered by Newbody 4 · 0⤊ 0⤋