Solve the system. y = 4x + 12 and y = -2x + 6
pick your method, substitution or elimination?
substitution looks ok since it only 2 equations and 2 variables and you are already solved for y.
4x + 12 = -2x + 6
-1,3 ?
Did you mean to say that, x = -1 and y = 3 @IllGiveYouMyAll ?
Yes, is that correct? @mathslover
x = -1 seems okay. But, y = 3 does not ...
How did you get to the above conclusion ?
I guessed. It's so confusing... Ugh.
See, is it clear to you , that : 4x + 12 = -2x + 6 ?
Well yes, but once i start solving it i seem to mess up.
No problem. We're here to help you with that. Okay. Let's start solving it. First of all, \(\boxed{4x + 12 = -2x + 6}\) Now, add \(\bf{2x}\) both sides. \(\implies 4x + 12 + \underline{2x} = -2x + \underline {2x} + 6\) [Notice the underlined "2x" there]
Now, can you tell me, what is : 4x + 12 + 2x ?
6x + 12?
Good :) . Now, similarly solve for : -2x + 2x + 6
6?
Very good ! So, you have now : 6x + 12 = -6 Can you solve for "x" from the above equation now?
3?
No , no! A little bit mistake. 6x + 12 = -6 Add 6 both sides 6x + 12 + 6 = -6 + 6 6x + 18 = 0 Subtract 18 both sides 6x + 18 - 18 = 0 - 18 6x = -18 Divide 6 both sides \(\cfrac{6x}{6} = \cfrac{-18}{6} \) x = -3
Oh, oh i see. I never know if when you multiply if it will be negative or positive.
Okay, now you have x = -3. You had in first, y = 4x + 12
Put the value of x = -3 in : y = 4x + 12 and solve for y
0?
Yes, but wait... a mistake there.
I did a mistake there . Sorry, : 6x + 12 = 6 [Not -6]
i.e. : \(6x + 12 = 6\) Sorry for the blunder.
Though, process will be exactly same. 6x + 12 = 6 Subtract 12 both sides 6x + 12 -12 = 6 - 12 6x = -6 x = -1
y = 4x + 12 [Given] y = 4(-1) + 12 Now solve for y.
So the answer is (-1,-3)?
No no! y = 4(-1) + 12 y = -4 + 12 y = ?
8!
Yep, good. So, y = 8 and the solution is : (-1,8)
Thank you SO much! c:
You're very welcome :)
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