Choose all of the system of equations which have no solution. x + 2y = 10 and 2y = 6 + 5x x = 8 − y and y = 6 + 5x 3x + 2y = 5 and 2y = 6 − 3x 3x = 2 − 4y and 4y = 7 − 3x

Third and fourth systems

Step-by-step explanation:

Let's check the first system:

x + 2y = 10

2y = 6 + 5x

Isolating x in the first equation, we have x = 10 - 2y. Applying that in the second equation we have:

2y = 6 + 5(10-2y)

2y = 6 + 50 - 10y

12y = 56

y = 4.6667

from the first equation, we have that x + 2*4.6667 = 10 -> x = 0.6667

So this system has a solution.

Checking now the second system:

x = 8 − y

y = 6 + 5x

using y from the second equation in the first equation, we have:

x = 8 - 6 - 5x

6x = 2

x = 0.3333

then, in the second equation:

y = 6 + 5*0.3333 = 7.6666

This system also has a solution

Third system:

3x + 2y = 5

2y = 6 − 3x

The second equation can be rewritten as:

3x + 2y = 6

We can see from both equation that the same expression (3x + 2y) has two values (5 and 6). So we can't solve this system.

Fourth system:

3x = 2 − 4y

4y = 7 − 3x

The first equation can be rewritten as 3x + 4y = 2

The second equation can be rewritten as 3x + 4y = 7

We can see from both equation that the same expression (3x + 4y) has two values (2 and 7). So we can't solve this system.

So, the systems that have no solution are the third and fourth systems

true? is that what you're asking or do you mean something else?

ian gone lie this got me confused

step-by-step explanation: