Question

2)(x - 2y=5,3x + 8y = 1;​

Answer 1

x-2y=5

3x+8y=1

x=5+2y

3(5+2y)+8y=1

x=5+2y

15+6y+8y=1

x=5+2y

15+14y=1

x=5+2y

14y=1-15

x=5+2y

14y=-14

x=5+2y

y=-14/14=-1

x=5+(-2)

y=-1

x=3

y=-1

Ответ: (3;-1)

Answer 2

To solve the system of equations $x - 2y = 5$ and $3x + 8y = 1$, you can use either the substitution method or the elimination method. I'll show you how to solve it using the elimination method:

First, let's manipulate the equations to eliminate one of the variables. Multiply the first equation by 3 so that the coefficients of $x$ in both equations will be the same:

$3(x - 2y) = 3(5)$
$3x - 6y = 15$

Now, you have the system:

$$\begin{align*} 3x - 6y &= 15 \\ 3x + 8y &= 1 \\ \end{align*}$$

Subtract the first equation from the second equation to eliminate $x$:

$$\begin{align*} (3x + 8y) - (3x - 6y) &= 1 - 15 \\ 14y &= -14 \\ y &= -1 \end{align*}$$

Now that you've found the value for $y$, substitute it back into the first equation to solve for $x$:

$$\begin{align*} x - 2(-1) &= 5 \\ x + 2 &= 5 \\ x &= 3 \end{align*}$$

Therefore, the solution to the system of equations is $x = 3$ and $y = -1$.