Can someone please explain how to work these problems? Which graph represents the solution? 1. y= -x+2 y=3x-1
id find the solution and see what quadrant its in ....
y= -x+2 y=3x-1 since y = y ... -x+2 = 3x-1 .. solve for x
once you know a value for x, plug it into either equation to find y
What? I am not following you.
do you agree that if: n = a, and n = b ; then a = b?
Yes.
well, we have stated here that: y = a and y = b , therefore a has to be equal to b they just gave equations instead ... y= -x+2 y= 3x-1 therefore -x+2 = 3x-1 this should be easily solvable for x now
But how do I solve it?
do you agree that if something is balanced; that in order to keep it balanced, you have to do the same thing to each side? for example: we know that 5 = 5 its balanced; to modify it, and keep it in balance, we would have to do the exact same thing to each side 5 = 5 5 - 2 = 5 - 2 (5-2)/7 = (5-2)/7 (5-2)/7 + x = (5-2)/7 + x do you agree that this has remained balanced thruout?
Yes.
Would that be how to solve the problem, or no?
then lets apply this principle to the balanced equations for our problem -x+2 = 3x - 1 ; lets add x to each side +x +x ------------ 0 + 2 = 4x - 1 what would you suggest we do next? keep in mind that we want x all alone
Simplify, 0+2 = 4x - 1.
ok 2 = 4x - 1 and next?
I don't know, ha. Ummm.. substitute?
lets work at getting rid of the -1 on the right ... i say we add 1, since -1+1 = 0 2 = 4x - 1 ; add 1 to each side +1 +1 ----------- 3 = 4x now we are one step away .... what would you suggest we do now?
Multiply 4 by itself, i don't know..
close, recall that anything DIVIDED BY itself is equal to 1, except 0/0 so, 4/4 = 1 , lets divide both sides by 4 3 = 4x ; divide by 4 /4 /4 ------- 3/4 = x and since they have decimal values on the graphs, x = .75 how many graphs have a crossing at x = .75?
The first one, has .75.
i agree, none of the others fit x = .75 so id go with that one :)
Thanks, for explaining this step by step with me I appreciate it. I'll give you a medal for best answer.
youre welcome, and good luck ;)
Thanks.
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