You are choosing between two plans at a discount warehouse. plan a offers an annual membership fee of $ 90and you pay 30 %of the manufacturer's recommended list price. plan b offers an annual membership fee of $ 40and you pay 80 %of the manufacturer's recommended list price. how many dollars of merchandise would you have to purchase in a year to pay the same amount under both plans? what will be the cost for each plan?

Step-by-step explanation:

We have 2 linear equations, and in both, the amount of merchandise you would have to purchase is "x", the unknown.  We are asked to find that value of x.

The first equation is

C(x) = .30x + 90, which says that the cost of this plan is a fixed $90, and you pay 30% of the manufacturer's cost, x.

The second equation is

C(x) = .80x + 40, which says that the cost of this plan is a fixed $40, and you pay 80% of the manufacturer's cost, x.

If we want to know when the cost of these 2 are equal to each other, we set the equations equal to each other and solve for x:

.3x + 90 = .8x + 40 so

-.5x  = -50 so

x = $100

The cost for each plan will be the same at this value of x, but we will plug in 100 for x in each just to make sure we did it right:

C(100) = .3(100) + 90

C(100) = 30 + 90

C(100) = 120  and

C(100) = .8(100) + 40

C(100) = 80 + 40

C(100) = 120

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0