SOLVE THE SYSTEM
(2D+3)(x)-(D+3)(y) = 5
(D+1)(x) - (2D+5)(y) = 7+4t
STEPS
1) eliminate x by mulitplying equation 1 with -(D+1) and equation 2 with -(2D+3)
After multiplying, what exactly happens to the D and t on the right side?
Does it become Dt or 0?
Equation 2 right side after multiplying: (7+4t)(-2D-3) = -14D -21 -8Dt -12t? or -14D-21??? OR IS IT SOMETHING ELSE?
(2D+3)(x)-(D+3)(y) = 5 ---times -(D+1) ---> -(D+1)(2D+3)(x)-(D+3)(-D-1)(y) = 5(-D-1)
-(D+1)(2D+3)(x) + (D^2 + 4D + 3)y = -5D - 5 **
(D+1)(x) - (2D+5)(y) = 7+4t ---times (2D+3) ----> (D+1)(2D+3)(x) - (2D+5)(2D+3)(y) = (7+4t)(2D+3)
(D+1)(2D+3)(x) + (-4D^2 - 16D - 15)y = 14D + 21 + 8Dt + 12t ****
add ** and ***
(-3D^2 - 12D + 12)y = 9D - 17 + 8Dt + 12t
y = (9D - 17 + 8Dt + 12t)/(-3D^2 - 12D + 12)
continue...
check my steps, I did not write this out first, very easy to make errors in typing