Question

Х-3у=5{ х+2у=10 х+2у=5{ х+6у=13 2х+5у=-12{ 2х+6у=4​

Answer 1

Ответ:

1) x=8; y=1;

2)x=1; y=2;

3)x=-46; y=16.

Пошаговое объяснение:

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Answer 2

It appears that you've given a system of linear equations. To solve this system of equations, you can use the method of substitution or elimination. Let's solve it using the elimination method:

Here are the equations:

1. $x - 3y = 5$
2. $x + 2y = 10$
3. $x + 2y = 5$
4. $x + 6y = 13$
5. $2x + 5y = -12$
6. $2x + 6y = 4$

First, let's simplify equations 2 and 3 by multiplying equation 3 by -1, so it becomes:

3. $-x - 2y = -5$

Now, you have:

1. $x - 3y = 5$
2. $x + 2y = 10$
3. $-x - 2y = -5$
4. $x + 6y = 13$
5. $2x + 5y = -12$
6. $2x + 6y = 4$

Now, you can use the elimination method to eliminate one variable. Let's add equations 1 and 3 together to eliminate $x$:

$x - 3y + (-x - 2y) = 5 - 5$

This simplifies to:

$-5y = 0$

Now, you can solve for $y$:

$y = 0$

Now that you have the value of $y$, you can substitute it into one of the original equations to solve for $x$. Let's use equation 2:

$x + 2(0) = 10$

This simplifies to:

$x = 10$

So, the solution to the system of equations is $x = 10$ and $y = 0$.