(a-d)/2c=b
4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)
Not sure about the original formula, but one that works is b = a/(2c + 1)
2b + 2c or 2(b + c)
(a + b)(b - 2c)
(b + 2c)(b - c)
(a-d)/2c=b
4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)
Not sure about the original formula, but one that works is b = a/(2c + 1)
2a + 2b = 26 2b + c = 22 ===> b = 11 - c/2 4c + 2b = 46 ===> b = 23 - 2c From the 2nd and 3rd equations: 11 - c/2 = 23 - 2c 22 - c = 46 - 4c 22 - 46 = -4c + c ===> -24 = -3c ===> c = 8 Substitute c=8 into the 3rd equation: b = 23 - 16 = 7 Substitute b=7 into the 1st equation: 2a + 14 = 26 2a = 12 ===> a = 6
That's going to depend on the values of 'a', 'b', 'c', and 'f'.
2b + 2c or 2(b + c)
2c + 4 - 3c = -9 + c + 5 -c + 4 = c - 4 -2c = -8 2c = 8 c = 4
1
I--2a+b=2c+2d II--2a+b=c+2d III--a+2b=3d * Here if we perform (I)-(II), we get, c=0 * Using the c we can prove b=0 using (I)-(III). * Now replace 'b' & 'c' as 0 in (I), then we get a=d. Email me at yash2008gates@gmail.com for any queries. I will be very thankful to receive your queries.
4c-7 = 2c+11 4c-2c = 11+7 2c = 18 c = 9
(a + b)(b - 2c)