5. How many solutions does each system of equations have?
a. 3y - 6x= 12
5y+10=15x

b 6y = 4x +9
12y-8x=18

C. Y+5x= 13
4y+17=-20x

1) The solution set is (1.2, -5.6)

2) It has no solution.

3) It has no solution.

Step-by-step explanation:

1) The system of equations are 3y - 6x= 12  and 5y+10=15x .

⇒ -6x + 3y = 12

Divide it by -3 on both sides,

2x -y = -4 (1)

⇒ 15x + 5y = -10

Divide it by 5 on both sides,

3x + y = -2 (2)

Adding (1) and (2),

2x -y = -4

3x + y = -2

5x      = -6

x = -6/5

x = 1.2

Substitute x=1.2 in (2),

3(1.2)+y = -2

y = -2 - 3.6

y = -5.6

The solution set is (x,y) = (1.2, -5.6)

2) The system of equations are 6y = 4x +9  and 12y-8x=18

4x -6y = -9 (1)

-8x +12y = 18 (2)

Multiply eq(1) by 2,

8x - 12y = -18.

Here, both the equations are equal and have opposite sign. Therefore, this system of equations cannot be solved for x and y value.

It has No solution.

3) The system of equations are y+5x= 13  and 4y+17=-20x

5x+y = 13 (1)

20x+4y = -17 (2)

Multiply eq(1) by 4,

20x+4y = 52

Here,  both the equations have same x and y terms and only constants are different. Therefore, this system of equations cannot be solved for x and y value.

It has No solution.

muyb ajo

step-by-step explanation:

jegosu

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40