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Wouldn't it have been much easier to make x=4-y and substitute that into xy=12 and xy=4? Then 4y-y^2=12 which becomes -y^2+4y-12=0 which becomes y^2-4y+12=0, and solve by quadratic equation. Similarly, 4y-y^2=4, which becomes - y^2+4y-4=0, which becomes y^2-4y+4=0, which becomes (y-2)(y-2)=0, which solves to y=2, therefore 2x=4, and x also =2.
Largo, pero muy explicado, aunque, confieso, una parte final no la capte en su definición. GUARDARÉ EL EJERCICIO Y LO VERÉ E INTENTARÉ, INTERPRETAR EL PASO A PASO. GRACIAS POR ENSEÑAR.
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THANKS PROFESOR !!!!, VERY INTERESTING !!!!
At first glance the answer is (2, 2) and of course there are some other complex solutions
I also found the 2nd solution instantly: x=y=2. But the 1st is difficult to found.
Thx a lot
Wouldn't it have been much easier to make x=4-y and substitute that into xy=12 and xy=4?
Then 4y-y^2=12 which becomes -y^2+4y-12=0 which becomes y^2-4y+12=0, and solve by quadratic equation.
Similarly, 4y-y^2=4, which becomes - y^2+4y-4=0, which becomes y^2-4y+4=0, which becomes (y-2)(y-2)=0, which solves to y=2, therefore 2x=4, and x also =2.
Largo, pero muy explicado, aunque, confieso, una parte final no la capte en su definición. GUARDARÉ EL EJERCICIO Y LO VERÉ E INTENTARÉ, INTERPRETAR EL PASO A PASO. GRACIAS POR ENSEÑAR.
Thanks 💯❤
Running time of video to problem solving can be reduced by avoiding writing of same thing again again. 24.21 mts can be reduced to less than 20 mts.
We just wrote EQ for equation.