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3 years ago

6 > 3p + 9 could u also show work

Mathematics

2 answers:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

p < -1

Step-by-step explanation:

Original equation:

6 > 3p + 9

Subtract 9 from both sides

The equation should look like this now:

-3 > 3p

Divide both sides by 3 (to separate the 3 from the p)

-1 > p

p < -1

Hope this helps :)

ddd [48]3 years ago

4 0

Answer:-1>p

Step-by-step explanation:

6>3p+9

-9 -9

-3>3p

Divide by 3 on each side

-1>p

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The value of x and y which satisfies the equation : -6x+7y=8

aleksandr82 [10.1K]

Answer:
x=1
y=2

Explanation:
first start with assigning a random value to a variable
lets say x=1
from there the equation becomes solveable
-6x1+7y=8
simplify
-6+7y=8
+6 to both sides
7y=14
divide both sides by 7
y=2

from here we can say that x=1, y=2, can satisfy the equation-6x+7y=8

hope this helps :)

8 0

3 years ago

What is the sum of all possible values of y if x is a positive integer, xy > 0, and 6x+2y=25?

SIZIF [17.4K]

Option E

The sum of all possible values of y is 20

Solution:

Given that,

6x + 2y = 25

Where, xy > 0

We have to find the sum of all possible values of y if x is a positive integer

We can use values of x = 1, 2, 3, 4

For x = 5 and above, y will be negative

Substitute x = 1 in given equation

6(1) + 2y = 25

6 + 2y = 25

2y = 25 - 6

2y = 19

Divide both the sides of equation by 2

$y = \frac{19}{2}$

Substitute x = 2 in given equation

6(2) + 2y = 25

2y = 25 - 12

2y = 13

Divide both the sides of equation by 2

$y = \frac{13}{2}$

Substitute x = 3 in given equation

6(3) + 2y = 25

2y = 25 - 18

2y = 7

Divide both the sides of equation by 2

$y = \frac{7}{2}$

Substitute x = 4 in given equation

6(4) + 2y = 25

24 + 2y = 25

2y = 25 - 24

2y = 1

Divide both the sides of equation by 2

$y = \frac{1}{2}$

Now add all the values of "y"

$\text{Sum of possible values of y } = \frac{19}{2} + \frac{13}{2} + \frac{7}{2} + \frac{1}{2}\\\\\text{Sum of possible values of y } = \frac{19+13+7+1}{2}\\\\\text{Sum of possible values of y } = \frac{40}{2}=20$