Question

1) (x + y = 12, 1 x - y = 2;2) J x + y = 19,1-x + y = 1;3) (2x + y = 10,[x - y =2.​

Answer 1

Ответ:

1) х=10 а у=2

2) непонял

3)х=4 а у=2

Answer 2

It looks like you've provided three sets of linear equations. If you're looking for solutions to these systems of equations, I can help you with that. Let's solve each system one by one:

1. \begin{align*} x + y &= 12 \\ x - y &= 2 \end{align*}

To solve this system, you can use the method of elimination or substitution. Let's use the elimination method. Add the two equations together:

$(x + y) + (x - y) = 12 + 2$ $2x = 14$ $x = 7$

Now, substitute the value of $x$ into one of the original equations (let's use the first equation):

$x + y = 12$ $7 + y = 12$ $y = 12 - 7$ $y = 5$

So, the solution to this system is $x = 7$ and $y = 5$.

2. \begin{align*} J + y &= 19 \\ 1 - x + y &= 1 \end{align*}

The second equation simplifies to $y = x$, so we can substitute this into the first equation:

$J + x = 19$

Without more information, we can't determine the exact values of $J$ and $x$, so the solution is not uniquely determined.

3. \begin{align*} 2x + y &= 10 \\ x - y &= 2 \end{align*}

Let's use the elimination method again. Add the two equations together:

$(2x + y) + (x - y) = 10 + 2$ $3x = 12$ $x = 4$

Substitute $x$ back into the second equation:

$x - y = 2$ $4 - y = 2$ $y = 4 - 2$ $y = 2$

The solution to this system is $x = 4$ and $y = 2$.

If you have any more equations or questions, feel free to ask!