What is the value of the y variable in the solution to the following system of equations? 3x + 2y = 6 2x + 3y = −6 A: 6 B: −3 C: 3 D: −6 *i got -3 but i want to make sure im right.
it might be right, how did you obtain it?
i dont get a -3
2(3x+2y)=2(6) 3(2x+3y)=3(-6)
well i did it some weird way , like my own way so thats why i wasn't sure if it was right
if we know the way you tried it, we can hone it better
6x + 4y = 12? 6x + 9y = -18? @jiteshmeghwal9
i have make one of the variable equal to gt the value of 'y' :)
equal but opposite is another strategy that eliminates an extra step out of that
i deleted my reply before u see
okay wait can you explain it to me again?
lol ;)
im soo confused....
it would be better to work off of the method you tried on your own .. imo
cool down friend amistre is here .. be happy and enjoy the learning @Rkb<3NH
yea but i got 6x = 2 at the end
im not concerned with the endl id like to see the process you did to correct it
ok wait
3x + 2y = 6 -3x -3x 2y = 6 - 3x 2x + 3y = −6 2x +(6-3X) 3y = −6 12x - 6x + 3y = -6 6x + 3y = -6 / 3 /3 6x = 2??
so your trying to do substitution
yea i tried elimation but to me it dident work out
you should start out by solving for an "x" to find the "y" in order to replace the "x"; we substitute it with its y equivalent does that make sense?
3x + 2y = 6 -2y -2y ----------- 3x = 6-2y x = (6-2y)/3
yes but why did you divide by 3?
because 3x does not equal x
3x/3 = x
3x = 6-2y /3 /3 ------------ x = (6-2y)/3
so i will be 2x + (6 -2y)/3 +3y = -6 12x + 4y/3 + 3y = -6 is that good so far?
4x +4/3y +3y = -6 i went wrong here, i dont knw
what to do
almost; lets clarify this "substitution" given that: x = (6-2y)/3 and 2x + 3y = -6 REPLACE (substitute) the equation of "x" in place of "x" \[2x + 3y = -6\] \[2\frac{(6-2y)}{3} + 3y = -6\] ^^^ we replaced "x"
12x - 4y/3 + 3y = -6 isn't that what i jut got because i cant divide 4/3
there shouldnt be any "x" left to play with in this equation
(12 - 4y)/3 + 3y = -6
4y - 4/3y + 3y = -6?
lets work this slowly (12 - 4y)/3 + 3y = -6\[\frac{12 - 4y}{3} + 3y = -6\]\[\frac{12}{3} - \frac{4y}{3} + 3y = -6\]\[4 - \frac{4}{3}y + 3y = -6\]-4 -4 -----------------\[- \frac{4}{3}y + 3y = -10\] good so far?
6x+4y=12 6x=12+4y x=12+4y/6
@jiteshmeghwal9 just a request to you : let amistre64 finish his explanation to the user .. please
k!
yes
and jite, your mathing; 6x+4y=12 6x 12+4y ; is off a little just saying :)
but 4/3y + 3y is 4.33333333...
Rkb, good. as long as its making sense to far, lets continue; without worrying about turning fractions into decimals .... \[- \frac{4}{3}y + 3y = -10\] \[\left(3-\frac{4}{3}\right)y = -10\] \[\left(\frac{9}{3}-\frac{4}{3}\right)y = -10\] \[\frac{5}{3}y = -10\] still good?
ok i get it now so it will be 3 * 5/3 = -10 * 3 5 = 30? good so far?
thats pretty good, dont forget the negative from that 10 tho 3 * 5/3 y = -10 * 3 5y = -30
oh okay: y = -6!!??(:
correct :) y = -6
cool math @amistre64 :)
thanx; elimination is quicker if you can work your head around it; but subbing is just as fine if your more comfortable with it
yes! i was going through elimination method ut @mathslover saide to me that we should see ur explanation first :)
thank you soo much @amistre64 (: And thank you tooo @jiteshmeghwal9 (: im glad you didn't just give me the answer you you helped me out(:
3x + 2y = 6 2x + 3y = −6 3x + 2y = 6 ; *2 2x + 3y = −6 ; *-3 6x + 4y = 12 -6x -9y = 18 ; combine them -------------- -5y = 30 /-5 /-5 ----------- y = -6
\[\Huge{\color{red}{you \space are \space welcome :) }}\]
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