Question

1239. 1) (6x – 8y -1,2 = 0,3х +5у - 0,2 = 0және(3x = 0,6+4y,15х = -1- 25у;Даю 40б ​

Answer 1

Ответ:

Пошаговое объяснение:

6x-8y-1,2=0 (разделили на 2) 3x=0,6+4y

3x+5y-0,2=0 равносильна (умножили на 5) 15x= -1-25y

10x-7y-1=0 (разделили на -10) 0,7y=x-0,1

-1,2y+6x+5=0 равносильна ( разделили на -1,2) y=5x+4 1/6

Answer 2

It seems like you've provided a system of equations and then a set of equations at the end, but the equations are written in a format that's a bit difficult to understand. I'll try to clarify and solve the system of equations for you:

1. $6x - 8y - 1.2 = 0$
2. $3x + 5y - 0.2 = 0$

And it seems like you have these two equations: 3) $3x = 0.6 + 4y$

4. $15x = -1 - 25y$

You'd like to find a solution for this system of equations.

First, let's simplify equations 3 and 4:

From equation 3, we can isolate x: $3x = 0.6 + 4y$ $x = 0.2 + 1.33y$

From equation 4, we can isolate x: $15x = -1 - 25y$ $x = -\frac{1}{15} - \frac{5}{3}y$

Now that we have expressions for x in terms of y from both equation 3 and 4, we can set them equal to each other:

$0.2 + 1.33y = -\frac{1}{15} - \frac{5}{3}y$

Now, we can solve for y:

$1.33y + \frac{5}{3}y = -\frac{1}{15} - 0.2$ $\frac{4.99}{3}y = -\frac{17}{15}$

Now, solve for y:

$y = \frac{-\frac{17}{15}}{\frac{4.99}{3}}$

Now, calculate y:

$y \approx \frac{-17}{15} \div \frac{4.99}{3} \approx -\frac{51}{4.99}$

Now that you have the value of y, you can substitute it back into one of the equations to find x. Let's use the equation $x = 0.2 + 1.33y$:

$x = 0.2 + 1.33 \cdot \left(-\frac{51}{4.99}\right)$

Now, calculate x:

$x \approx 0.2 - \frac{1.33 \cdot 51}{4.99} \approx -\frac{156.83}{4.99}$

So, the approximate solution to the system of equations is:

$x \approx -\frac{156.83}{4.99}$ $y \approx -\frac{51}{4.99}$