You can put this solution on YOUR website! (500x+100y=325)
(400x+150y=304)
Simplify each equation so we can use elimination
divide the 2nd equation by 40 and the 1st equation by 50
10x + 3.75y = 7.6
10x + 2.0y = 6.5
-------------------subtraction eliminates x, find y
0 + 1.75y = 1.1
y = 1.1/1.75
y = .62857
Use the first equation to find x
500x = 100(.62857) = 325
500x = 325 - 62.857
500x = 262.143
x = 262.143/500
x = .52429
:
:
See if this checks out in the 2nd equation
400(.52429) + 150(.62857) =
209.7144 + 94.2855 = 303.9999 ~ 304
You can put this solution on YOUR website! ....eq.1
....eq.2
---------------------------I guess you are given a system of linear equations to solve
start with
....eq.1, and sove for .....eq.1a
do same with
....eq.2
..........eq.2a
from eq.1a and eq.2a we have:
...solve for
This is a perfect example of what NOT TO DO when solving a system with large numbers and possibly fractional solutions.
One of these 2 people is WRONG, as it's most likely attributable to the fact that both converted to decimals, which is TOTALLY UNNECESSARY, and as in most cases and this,
such "strategy," if you can call it that, can lead to possible confusion, and INCORRECT answers. This is the way it should be done:
500x + 100y = 325 ------ eq (i)
400x + 150y = 304 ------ eq (ii)
100x - 50y = 21 -------- Subtracting eq (ii) from eq (i) ------ eq (iii)
100x + 20y = 65 -------- Dividing eq (i) by 5 ------ eq (iv)
70y = 44 -------- Subtracting eq (iii) from eq (iv)
Since it's too complex to substitute (fraction that doesn't cancel nicely) into any of the 2 ORIGINAL equations - and could likely result in error(s) - we will ELIMINATE y as follows:
300x - 150y = 63 --- Multiplying eq (iii) by 3 ------- eq (v)
700x = 367 --------- Adding eqs (v) and (ii)
You don't have to go further and convert the fractions to decimals. It's TOTALLY UNNECESSARY; plus, you were NOT asked to do so.
You can put this solution on YOUR website! .
With so ugly numbers, I think that the most straightforward way is to use determinants ( = the Cramer's rule for 2x2-system).
