Darren wins a coupon for 4$ off the lunch special for each of 5 days. he pays $75 for his 5 lunch specials. write and solve an equation to find the original price p for one lunch special

Value of each coupon = $4.

Total number of lunches taken = 5.

Let us assume original price for one lunch special = p.

Total amount paid in all 5 lunches taken = $75.

We can setup an equation now,

Number of lunches( cost of each lunch - discount value of each coupon) = $75

Plugging values,

5 ( p - 4) = 75.

This is the required equation to find the original price p for one lunch special.

Distributing 5 over (p -4), we get 5*p +5*-4 =  5p -20.

5p -20 =75.

Adding 20 on both sides.

5p -20+20 =75+20

5p = 95.

Dividing both sides by 5, we get

5p/5 = 95/5

p = 19.

Therefore, the original price is $19 for one lunch special.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."