Question

1) x+y=12 x-y=2 2) x+y=19 -x+y=1 3)2x+y=10 x-y=2

Answer 1

1) x+y=12; x-y=2

x+y=12; x=2+y

2+y+y=12

2+2y=12

2y=10

y=5

x+5=12

x=7

2)x+y=19; -x+y=1

x+y=19; y=1+x

x+1+x=19

2x=18

x=9

9+y=19

y=10

3)2x+y=10; x-y=2

2x+y=10; x=2+y

2(2+y)+y=10

4+2y+y=10

4+3y=10

3y=6

y=2

2x+2=10

2x=8

x=4

Ответы: 1)х=7; у=5, 2)х=9;у=10, 3)х=4; у=2.

Answer 2

Let's solve each system of equations:

1) $x + y = 12$ and $x - y = 2$:

To solve this system, you can use the addition method (elimination method):

Add the two equations:

$$\begin{align*} (x + y) + (x - y) &= 12 + 2 \\ 2x &= 14 \\ x &= 7 \end{align*}$$

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

$$\begin{align*} 7 + y &= 12 \\ y &= 5 \end{align*}$$

So the solution is $x = 7$ and $y = 5$.

2) $x + y = 19$ and $-x + y = 1$:

Again, you can use the addition method. Add the two equations:

$$\begin{align*} (x + y) + (-x + y) &= 19 + 1 \\ 2y &= 20 \\ y &= 10 \end{align*}$$

Now substitute $y$ back into one of the original equations, let's use the first one:

$$\begin{align*} x + 10 &= 19 \\ x &= 9 \end{align*}$$

So the solution is $x = 9$ and $y = 10$.

3) $2x + y = 10$ and $x - y = 2$:

For this system, let's use the substitution method. Solve the second equation for $x$:

$$\begin{align*} x - y &= 2 \\ x &= y + 2 \end{align*}$$

Now substitute this expression for $x$ into the first equation:

$$\begin{align*} 2(y + 2) + y &= 10 \\ 2y + 4 + y &= 10 \\ 3y &= 6 \\ y &= 2 \end{align*}$$

Now substitute the value of $y$ back into the expression for $x$:

$$\begin{align*} x &= y + 2 \\ x &= 2 + 2 \\ x &= 4 \end{align*}$$

So the solution is $x = 4$ and $y = 2$.