You can almost tell how this will turn out just by looking. If you divide all terms in the first equation by 2, the result is a copy of the third equation. Using the two identical equations to eliminate a variable cancels ALL terms to zero, leaving you with a single equation in 3 variables, which has an infinite number of solutions.
Your second analysis of why there are infinitely many solutions is not quite correct. You used equations 2 and 3 to eliminate y, but - x - z = 7 alone does not imply infinitely many solutions. If this system had a solution, and you were going to solve it using elimination, you could eliminate y using equations 2 and 3, but you would also have to eliminate y using a different pair of equations. If you were unlucky enough to use equations 1 and 2, you would get a second equation in x and z alone. You would then take your first equation (which you call equation B) and this second equation and solve a 2x2 system. THAT process would yield a 0 = 0 case, and hence infinitely many solutions.
y really is 6 , but z and x got infinite solution , x+z = -7 , z could be the -7 and x could be 0 , any number is possible as long x = -z - 7 and z = -x - 7
You can almost tell how this will turn out just by looking.
If you divide all terms in the first equation by 2, the result is a copy of the third equation. Using the two identical equations to eliminate a variable cancels ALL terms to zero, leaving you with a single equation in 3 variables, which has an infinite number of solutions.
Well, you just saved my GPA by a full 2 points lol
You don't actually show how to solve this...
Your second analysis of why there are infinitely many solutions is not quite correct. You used equations 2 and 3 to eliminate y, but - x - z = 7 alone does not imply infinitely many solutions. If this system had a solution, and you were going to solve it using elimination, you could eliminate y using equations 2 and 3, but you would also have to eliminate y using a different pair of equations. If you were unlucky enough to use equations 1 and 2, you would get a second equation in x and z alone. You would then take your first equation (which you call equation B) and this second equation and solve a 2x2 system. THAT process would yield a 0 = 0 case, and hence infinitely many solutions.
Please provide playlist of matrix
Thats not really solving the system though thats just saying that it has infinite solutions and its dependent
Yeah, but in hs algebra 2 they really just want you to say “there’s infinitely many solutions” and that’s that.
Scared of how you solve them though-
@@Charlie-hv3dhbruh this is on the first test of the year and im only a sophomore so idk if i should drop the class or no
@@Charlie-hv3dh nah for us we actually have to solve for it [using one variable so for this i think it's (x, 5-x, -6)]
ahh alright @@nancy-zl8tl
omg a fellow swiftie@@nancy-zl8tl
You could solve it easily with Cramer's Rule
I got Y=6; z=0; x= -7
do not see a mistake in my end
y really is 6 , but z and x got infinite solution , x+z = -7 , z could be the -7 and x could be 0 , any number is possible as long x = -z - 7 and z = -x - 7
I dont know if i am right
I do not see a mistake on my end
I got the vlue of x y and z
By multiplying the 1st equation by 3 and the 2nd Equation I multiplied it by -2
there are multiple ways to solve
Ah.. thanks
Not helping
Well, that was a waste of time.
@@KW-gb9cd Yes if you already knew this. Otherwise you would've been stumped when solving it for the first time