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Thank you sir for your cooperation.

Nhi hua

bhai kuch bol to

🙏🙏

🙇🙇🙇

I got it right first try 😎 not a hard problem

2

By inspection 4x3=12 and 2x9=18, therefore x=2 and y=1. Took about 1/5 of the time

طريقة صحيحة

Please which software are you using to edit your thumbnail video

Clearly done, but no need to work out both scenarios. From (1) / (2) you got x - y = 1 Just plug it back into (1) 2^(1-y) x 3^y = 12 so 2^y x 3^y = 6 6^y = 6 => y = 1, so x = 2

Bro really 💀

Why divide by 2? Instead I multiplied it first by 2, then by 4 (so times 8 overall) thus avoiding fractions in my working until the last step.

Ok good one

Ok good one

where is 4 coming from

It. Can. Do. Medilturm. Fraction. For. X.

Nice explanation

Well solved

Division 1-i -1

Multiplication 34-13i 5i-3

Subtractions 6+4i _2+12i

5+2i 4+13i

Easily factored instead of completing the square: (2x - 3)(x + 3) = 0, so x = 3/2 & -3.

X= 2 y=1

0 , -1 , 3

Nice video

X=3 or -1

Pythagorean theorem

After x=2 at 0.48 by elimination, substitute for x into either equation and solve for y. No need for a second elimination process. 🇦🇺

2*5√2 = √2x 10√2=√2x 100*2=2x 200=2x x=100

3x^2+3x+4x+4=0 3x(x+1)+4(x+1) (3x+4)(x+1) x=-4/3 dan x=-1

😢 "PromoSM"

Wonderfully done!

THANKS PROFESOR !!!, VERY INTERESTING !!!

ANYTHING 'to the power 0' = 1 So x = 0

5x-y=4 2x+y=10 5x+2x-y+y=4+10 7x=14 x=2 5(2)-y=4 10-y=4 -y=-6 y=6 2(2)+y=10 4+y=10 y=6 (x, y)=(2, 6) ❤

Kam se kam solution toh pura dete

Ye kya tha😂

_/x+_/(x+7)=7 _/x-7=-_/(x+7) x+49-14_/x=x+7 49-14_/x=7 -14_/x=-42 _/x=3 x=9 ❤

7^-1/3

x=0

X=0

Good explanation. Your students must be happy. If you are a teacher, that is.

x = 0. Not much to solve.

A more general solution would be nice.

LOL. Got me with an easy one.

You can solve this in half a second if you remember that anything to the zero-power equals 1. Solving it the long(er) way, take the log of both sides: x log(5) = log(1) x = log(1)/log(5) log(1) = 0, so x = 0. You have to worry a little bit that there may be more than one solution. Exponential functions sometimes get mysterious between 0 and 1 and finding intersection(s) of two exponential functions can get complicated and have multiple solutions. With any constant to the x-power and you want to find where it equals a constant, you won't have that problem. When x is decreasing, it goes asymptotic with the x-axis. When x is increasing, it keeps going up and up and it's not turning back toward the x-axis at any point (you can prove all this with derivatives). So, the function y=5^x crosses y=1 only once. x=0, you're done. If you're pressed for time on a test and don't have time to ponder all these nuances, take the easy solution x=0 and move on, next question. Getting half-credit on a question is better than not finishing the test.

Any non-zero quantity raised to the power of 0 is defined to be 1; therefore x=0 is a solution to this problem.