ok here is what i got: 8x+5y=184 (1) x-y=3 (2) from equation (1) 8x+5y=184 => 8x=184-5y x=184-5y/8 call this equation (3) substitute equation (3) in equation (2) 184-5y/8-y =3 (184/8)-(5y/8)-y=3 (184/8)-(13y/8)=3 (184/8)-3=13y/8 160/8=13y/8 20=13y/8 [cross multiply] 20(8)=1...
This is the substitution method y-2x=8 (1) 2y+x=1 (2) From equation (1) y-2x=8 would now be y=8+2x you can call this equation (3) so now, we substitute equation (3) in equation (2) 2(8+2x)+x=1 16+4x+x=1 16+5x=1 5x=1-16 5x=-15 x=-15/5 x=-3 for x=-3, substitute it in equation (3...
Well these would be two equation so: y=6-x (1) y=x+2 (2) so, substitute equation (1) y=6-x into equation (2) therefore, 6-x=x+2 6-2=x+x 4=2x 4/2=x 2=x for x=2, substitute x in equation (1) y=6-2 y=4 x=2 , y=4
how do i evaluated these limits: lim x --> 3 ((square root of x^2 +16)- 5)/(x^2 - 3x) lim x--> to infinity (2+ 100/x)
oh don't worry drwls, i now understand. thanks!
sorry i mean the top looks like a quadratic do i break it down or just leave it as it is and just plug in x=5 into the entire equation?
so drwls do i have to change the top into something similar or just plug in x=5 into the entire equation.
ok lemme write it over then: How do i find the lim x-->5 (x-3-2x^2)/(1+3x)
how do i evaluate this limit: lim x tends to 5 x-3-2x^2/1+3x
Hey thanks a lot Reiny. I got it..