Determine whether the systems have one solution, no solution, or infinitely many solutions.
3x - 2y = 3; 6x - 4y = 1
3x - 5y = 8,5x -- 3y = 2
3x + 2y = 8; 4x + 3y = 1
3x - y = 3; 2x - 4y = 2
3x - 4y = 2; 6x - y = 1

First equation: no solution

Second equation: one solution

Third equation: one solution

Fourth equation: one solution

Fifth equation: one solution

Step-by-step explanation:

First system: 3x - 2y = 3; 6x - 4y = 1

Multiplying the first equation by two, we have:

6x - 4y = 6

As this equation and the second equation of the system have the same coefficients to x and y, but a different independent value (1 and 6), the system has no solution

Second system: 3x - 5y = 8; 5x - 3y = 2

From the first equation: x = (8 + 5y)/3

Using this value of x in the second equation, we have:

(40 + 25y)/3 - 3y = 2

40 + 25y - 9y = 6

16y = 34

y = 2.125

x = (8 + 5*2.125)/3 = 6.2083

The system has one solution

Third system: 3x + 2y = 8; 4x + 3y = 1

From the first equation: x = (8 - 2y)/3

Using this value of x in the second equation, we have:

(32 - 8y)/3 +3y = 1

32 - 8y + 9y = 3

y = -29

x = (8 - 2*(-29))/3 = 22

The system has one solution

Fourth system: 3x - y = 3; 2x - 4y = 2

From the first equation: y = 3x - 3

Using this value of y in the second equation, we have:

2x - 12x + 12 = 2

10x = 10

x = 1

y = 3 - 3 = 0

The system has one solution

Fifth system: 3x - 4y = 2; 6x - y = 1

From the first equation: 3x = 2 + 4y

Using this value of 3x in the second equation, we have:

4 + 8y - y = 1

7y = -3

y = -0.4286

x = (2 + 4y) / 3 = (2 + 4*(-0.4286)) / 3 = 0.0952

The system has one solution

One solution:

3x- 5y = 8; 5x - 3y = 2

3x + 2y = 8; 4x +3y = 1

No solution:

3x - 2y = 3; 6x - 4y = 1

3x - 4y = 2; 6x - 8y = 1

Infinitely solutions:

3 x - 6y = 3; 2x - 4y = 2

Step-by-step explanation:

Just got it right on edge 2020

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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