Can someone teach me how to solve this? 3(y-2)-y=2(y+3)
ok (A) x² - y² = 3, (B) x + y = 2 => x = 2 - y Substituting (B) into (A), (2 - y)² - y² = 3 y² - 4y + 4 - y² = 3 - 4y = - 1 4y = 1 y = 1/4 Substituting back into (B), x = 2 - 1/4 = 7/4 (7/4, 1/4)
did this help u @Annetta_Martin
That actually confused me even more....
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omg lol
Multiply it out. Bring y terms to the left and constant terms to the right. Simplify and solve for y.
ok Rearrange equation 2 to get y=2-x Substitute this into equation 1 x^2-(2-x)^2 = 3 x^2-(4-4x+x^2)=3 x^2-x^2+4x-4=3 4x-4=3 4x=7 x=7/4 Substitute this value of x into equation 2 to work out y 7/4 + y=2 y=2-(7/4) = 1/4 Check using equation 1 (7/4)^2-(1/4)^2=3 3.0625-0.0625=3 And there you have it ;) x=7/4=1.75 y=1/4=0.25
is this better??????????????
hello is this better
realmeeee your explanation seems fine if thats confusing then you dont understnad math
3(y-2)-y=2(y+3) 2y-6= 2y+6 I don't even get a solution? lol
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i do understand matth
X_X
3(y-2)-y=2(y+3) 3y - 6 - y = 2y + 6 2y - 6 = 2y + 6 add 6 to both sides: 2y = 2y + 12 subtract 2y from both sides: 0 = 12 ==> inconsistent. Therefore, there no solution to this problem.
not you realmee im saying the person reading your explanation doesnt understand math i thought your explanation was good
That makes more sense, thanks @ranga
you are welcome.
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@sweetburger Math is my weak subject, but to go around and point out people's flaws in such a manner, it just shows that you are poorly educated and lack manners. What a shame... Oh well! :)
that should be my medal
i am not well educated you are right @Anneta_Martin
\[3(y-2)-y=2(y+3)\] distributive property gives \[3y-6-y=2y+6\] combine like terms on the left and get \[2y-6=2y+6\] and then note that it is not possible for \(2y-6\) to also be equal to \(2y+6\) so there is no solution
@TheRealMeeeee It's a virtual medal dude... your satisfaction should be her understanding it. :P
hope it is clear that you cannot add 6 to a number, subtract 6 from that same number, and get an equality
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