Question

2 (2x+1)-(2y-3)=3 -8x+3(y-2)=-4​

Answer 1

$\left \{ {{2(2x+1)-(2y-3)=3} \atop {-8x+3(y-2)=-4}} \right. \\\\\left \{ {{4x+2-2y+3=3} \atop {-8x+3y-6=-4}} \right.\\\\\left \{ {{4x-2y=-2}|*2 \atop {-8x+3y=2}} \right. \\\\+\left \{ {{8x-4y=-4} \atop {-8x+3y=2}} \right. \\-------\\-y=-2\\\\y=2\\\\4x=2y-2=2*2-2=2\\\\x=0,5\\\\Otvet:\boxed{(0,5;2)}$

Answer 2

To solve the system of equations, we will start by simplifying each equation and then solving for one variable in terms of the other. Once we have one variable's value, we can substitute it back into the other equation to find the value of the other variable.

Given system of equations:

1. $2(2x+1) - (2y-3) = 3$
2. $-8x + 3(y-2) = -4$

Let's simplify these equations step by step:

Equation 1: $4x + 2 - 2y + 3 = 3$
$4x - 2y + 5 = 3$
$4x - 2y = -2$
$2x - y = -1$ (Divided both sides by 2)

Equation 2: $-8x + 3y - 6 = -4$
$-8x + 3y = 2$

Now we have a simplified system of equations:

1. $2x - y = -1$
2. $-8x + 3y = 2$

We can use the first equation to solve for $y$ in terms of $x$:

$2x - y = -1$
$y = 2x + 1$

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

$-8x + 3(2x + 1) = 2$
$-8x + 6x + 3 = 2$
$-2x + 3 = 2$
$-2x = -1$
$x = \frac{1}{2}$

Now that we have the value of $x$, we can substitute it back into the expression for $y$:

$y = 2x + 1$
$y = 2 \cdot \frac{1}{2} + 1$
$y = 1 + 1$
$y = 2$

So, the solution to the system of equations is $x = \frac{1}{2}$ and $y = 2$.