Answer:
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
