Mr. lee asks a student to solve the following system of linear equations. {-28=-40q-14p 10q+4=2p which ordered pair (p, q) is a solution to the system? (-7/5,-2) (-2,-7/5) (0, 2) (2, 0)

p = 2 , q = 0

Step-by-step explanation:

Solve the following system:

{-28 = -14 p - 40 q | (equation 1)

10 q + 4 = 2 p | (equation 2)

Express the system in standard form:

{14 p + 40 q = 28 | (equation 1)

-(2 p) + 10 q = -4 | (equation 2)

Add 1/7 × (equation 1) to equation 2:

{14 p + 40 q = 28 | (equation 1)

0 p+(110 q)/7 = 0 | (equation 2)

Divide equation 1 by 2:

{7 p + 20 q = 14 | (equation 1)

0 p+(110 q)/7 = 0 | (equation 2)

Multiply equation 2 by 7/110:

{7 p + 20 q = 14 | (equation 1)

0 p+q = 0 | (equation 2)

Subtract 20 × (equation 2) from equation 1:

{7 p+0 q = 14 | (equation 1)

0 p+q = 0 | (equation 2)

Divide equation 1 by 7:

{p+0 q = 2 | (equation 1)

0 p+q = 0 | (equation 2)

Collect results:

 {p = 2 , q = 0

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.