Samuel wants to buy at least twice as many dvds as cds. cds sell for $7 each, and dvds sell for $12 each. samuel can spend a maximum of $50. let c represent the number of cds, and let d represent the number of dvds. which system of inequalities can be used to find the number of cds and dvds samuel can buy?

The inequalities that will be used are:

d ≥ 2c

7c+12d ≤ 50

Step-by-step explanation:

Let:

c = number of CDs purchased

d = number of DVDs purchased.

It is given that:

Samuel wants to buy at least twice as many DVDs as CDs.(1)

CDs sell for $7 each, and DVDs sell for $12 each.

Samuel can spend a maximum of $50.(2)

Let c represent the number of CDs, and let d represent the number of DVDs.

Hence, we have from first statement that:

d ≥ 2c.

From the second statement we have:

7c+12d ≤ 50

Hence, the inequalities used are:

d ≥ 2c

7c+12d ≤ 50.

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

df =

both have same length