For what value of k does the equation 6(x + 1) + 2 = 3(k5x + 1) + 3 have no solution?

k = (6/15)

Step-by-step explanation:

The equation is:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

To have no solutions, we need to have something like:

x + 7 = x + 4

where we can remove x in both sides and end with

7 = 4

So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.

First, let's try to simplify our equation:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

6*x + 6 + 2 = 3*k*5*x + 3*1 + 3

6*x + 8 = 15*k*x + 6

if 15*k = 6, then the system clerly has no solution.

then:

k = 6/15

then we get:

6*x + 8 = (6/15)*15*x + 6

6*x + 8 = 6*x + 6

8 = 6

The system has no solutions.

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -