Question

X+y=2500, 21x-0,08y=38​

Answer 1

Ответ:

y = 50

x = 200

Пошаговое объяснение:

x+y=250

0, 21x-0,08y=38​

x = 250 - y

0, 21(250 - y) - 0,08y = 38

52,5 - 0,21y - 0,08y = 38

0.29y = 14,5

y = 14,5/0.29

y = 50

x = 250 - y

x = 250 - 50

x = 200

Answer 2

It seems you've provided a system of linear equations:

1. $x + y = 2500$
2. $21x - 0.08y = 38$

To solve for the values of $x$ and $y$, we can use various methods such as substitution, elimination, or matrix methods. I'll use the elimination method here:

First, let's multiply the second equation by 100 to get rid of the decimal:

$2100x - 8y = 3800$

Now, we can subtract the first equation from the modified second equation to eliminate $y$:

$(2100x - 8y) - (x + y) = 3800 - 2500$ $2100x - 8y - x - y = 1300$ $2099x - 9y = 1300$

Now, we can solve this equation for $y$:

$9y = 2099x - 1300$ $y = \frac{2099x - 1300}{9}$

Now we can substitute this expression for $y$ into the first equation:

$x + \frac{2099x - 1300}{9} = 2500$

Solve for $x$:

$\frac{9x + 2099x - 1300}{9} = 2500$ $9x + 2099x - 1300 = 22500$ $2108x - 1300 = 22500$ $2108x = 23800$ $x = \frac{23800}{2108}$ $x \approx 11.29$

Now that we have $x$, we can plug it back into the equation for $y$:

$y = \frac{2099x - 1300}{9}$ $y = \frac{2099 \cdot 11.29 - 1300}{9}$ $y \approx 2488.56$

So, the approximate solutions are $x \approx 11.29$ and $y \approx 2488.56$. Keep in mind that these are rounded values due to the approximations involved in calculations.