SOLUTION: How do you solve when decimals are involved? 3x-2.5y=7.125 2.5x-3y=7.3125

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Question 3392: How do you solve when decimals are involved?
3x-2.5y=7.125
2.5x-3y=7.3125

Found 2 solutions by drglass, gsmani_iyer:
Answer by drglass(89) About Me  (Show Source):
You can put this solution on YOUR website!
The number of real solutions to a quadratic equation is either 0, 1 or 2 depending on certain criteria. The easiest way to find the number is to look a something called the discriminant = b%5E2+-+4ac. The values for a, b and c are found in the equation itself. A quadratic equation always has the form ax%5E2+%2B+bx+%2B+c+=+0. In your example, a = 2, b = 4 and c = -4. The discriminant in the case would be 4%5E2+-+4%2A%282%29%2A%28-4%29 since this value is greater than 0, we know there will be two real valued solutions. In fact they would be %28-4+%2B+sqrt%284%5E2+-+4%2A%282%29%2A%28-4%29%29%29%2F%282%2A4%29 and %28-4+-+sqrt%284%5E2+-+4%2A%282%29%2A%28-4%29%29%29%2F%282%2A4%29. If the discriminant were zero, it would have one real solution and if it were less than zero, it would have no real valued solutions, but it would have two complex solutions.

In short, this equation has two solutions.

Answer by gsmani_iyer(201) About Me  (Show Source):
You can put this solution on YOUR website!

3x - 2.5y = 7.125 .... (1)
2.5x - 3y = 7.3125 .... (2)
by multiplying eq.(1) by 2.5, we get
7.5x - 6.25y = 17.8125 ... (3)
similarly eq.(2)*3, we get
7.5x - 9y = 21.9375 .... (4)
now, eq.(4)-(3), we get
-2.75y = 4.125
So y = 4.125%2F-2.75 = -1.5
replacing the value of this y in eq.(1)
3x -2.5%2A%28-1.5%29 = 7.125
= 3x - (-3.75) = 7.125
= 3x + 3.75 = 7.125
= 3x = 7.125 - 3.75 = 3.375
So x = 3.375%2F3 = 1.125

x = 1.125; y = -1.5 Answer.

gsm