I swear Kelton's Algebra is going to be the death of me in a few minutes.. Kelton needs some help on this:
a-2b = 2 a-2b+2b=2+2b a=2+2b a+b = 2 (b+3) 2+2b+b=2(b+3) 2+3b=2b+6 2+3b-3b=2b-3b+6 2-6=-b+6-6 -4=-b -1 -1 4=b a-2b=2 a-2(4)=2 a-8=2 a-8+8=2+8 a=10 does that help cause if a-2b=2 then 10-8=2 and 10+4=2(4+3) 14=14 done
Doing the linear combination method you basically try to eliminate a variable using two equations so: 1. a-2b=2 a+b=2(b+3) 2. So you are going to want to get a top variable to equal a bottom variable so I decided to eliminate the a variable and to do that you multiply the bottom equation by -1 a-2b = 2 -a-b=-2(b+3) 3. So then you add together the top and bottom equation so you should get: -3b=-2b-4 4. Simplify that down to: -b=-4 5. So b=4 6. And after that you can just plug in 4 for b and solve for a. Or you can solve it the linear combination method. Either way you get a=10
So what you want to do is isolate for one unknown so you end up with a=whatever and that there are no a's on the other side of the = sign. then you can plop the "whatever" part of the equation into where ever it saya a...because you know what a equals. So a-2b = 2 isolate the a by adding 2b to each side of the equation in order to keep it even on both sides a - 2b (+2b) = 2 (+2b) reduce the left side (ie the -2b and the +2b cancel each other out) so now you have: a=2 + 2b Now you know what a equals. Use that in the second equation and substitute (2+2b) for a (i'm using brackets to try to show you the substitutions) starting equation a+b = 2 (b+3) substitute in a=2+2b (2+2b)+b= 2 (b+3) collect like variables 2+3b=2(b+3) here you are multiplying 2 across the brackets 2+3b=2xb + 2x3 do multiplication 2+3b= 2b +6 (subtract 2b from both sides to keep it even) 2+ 3b-2b=2b+6-2b collect like variables and do addition/subtraction 2+ 1b=6 (subtract 2 from both sides to isolate the varriables) 2+ 1b-2 =6-2 do addition/subtraction 1b=4 b=4 you can check by subsituting b=4 back into the equation where you had only bs (2+2b)+b= 2 (b+3) czardoust shows the check. hth Karen
Thank you to all of you for your help..I had Kelton look at this thread and he then said "oh yeah" Good thing he knows what you all wrote down, because once I saw the first a= my brain shut down
LMAO....thats like me and my oldest daughter with her Algebra 2 & Trig homework Tangents, sins and co-sins! YAH right.
By that time I will be a walking zombie and he will just have to rely on tutors or by asking the professor for help.
Trig's a snap. You just have to remember Chief Sohcahtoa. Sin = opposite over hypotenus Cosin = adjacent over hypotenus Tangent = opposite over adjacent
My 18 year old starts tutoring a college sophomore tomorrow for her Algebra class. It's good practice for him since he wants to teach HS math. I can find out if he can squeeze you in.