Question

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

Answer 1

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