Finding the value of x+y+z|Learn how to solve a system with three unknowns

[xZHRA]

I JUST CALCULATED THAT IN MY HEAD IN 10 SECONDS, IM TOO SMART

[Skank_and_Gutterboy]

If you're going to make any restrictions x, y, z>0, put that in the problem statement up-front. I paused the video one-second in and solved it without using that constraint.

[SuperYoonHo]

because
x^2y^2z^2=24*48^72
of 6*3*2=48*6=288
since we are given three equations we get z=+-288/24 y=+-288/48 and lastly x=+-288/72
so z=z/1 y=z/2 and x=z/3 so x+y+z=z/1+z/2+z/3=11z/6
which turns out to br +-22
i did it this way i was right.
when taking the sqrt always consider the +or- sign in front when only the root sign it is just the pos root
but in this que ther ws no given sqrt sign so we must consider +or- and it makes sence ina way
pos*pos=pos
neg*neg=pos
so it workes out fine. do not worrie even great mathematicians make mistakes

[carlinoiavarone]

x= 24/ y thennwe can substitute the other terms of equations

[olivierheiderich]

extremely complicated way to calculate. You can easily see that z=2y, y=3/2*x and therefore z=3x, thus 3x²=48 -> x=4 -> x+y+z=11/2*x=22

[vijaysingbundhoo]

From the given equations x equals 2k... y equals 3k...and z equals 6k for some k. Easily seen that x and y and z equals 4and 6 and 12 respectively and the
result follows...

[prabhudasmandal]

Another way.xz/xy=48/24 or z/y=2.now, z/y×yz=2×72 or z^2=144 or z=12. Now xz=48 and z=12.so, x=4, again, yz=72 and z=12, so, y=72/12=6.hence, x+y+z=4+6+12=22.(--ve values of x, y, z are rejected since they are less than 0).

[Skank_and_Gutterboy]

You really did this the hard way. I solved the 2nd equation for x: x=48z. I solved the 3rd equation for y: y=72/z. Then I substituted both of these into the first equation so that you get an equation just in terms of z: (48/z)(72/z)=24, which rearranges z^2=144. Therefore, z=(12, -12). Then the other variables are easily solved, y=(6, -6) and x=(4, -4). A check of (x, y, z) = (4, 6, 12), (-4, -6, -12) shows that both sets satisfy the original system. So x+y+z=22 or -22.

I know, the instructor called out the x, y, z>0 but he didn't call that out soon enough, I paused the video 1-second in and solved it.

[umtelespectadorqualquer]

x×y+z×y=(x+z)×y
x×y-z×y=(x-z)×y
x÷y+z÷y=(x+z)÷y
x÷y-z÷y=(x-z)÷y
x+y÷z =(n+y)÷z n=x×z
x+y×z =(n+y)×z n=x÷z
x+y ×raiz de x =(raiz de x +y)raiz de x _o y não esta sob raiz_
x+y ÷raiz de x =(raiz de x +y)÷raiz de x
_o y não esta sob raiz_