Four cookies and three cupcakes code $13.25. Five cookies and two cupcakes cost $11.75. Give the system of equations that could
be used to find the cost of one cookie and the cost of one cupcake. Solve the system.
1 answer:
Answer:
- cookie: $1.25
- cupcake: $2.75
Step-by-step explanation:
Let x and y represent the cost of a cookie and a cupcake, respectively.
4x +3y = 13.25
5x +2y = 11.75 . . . . . . the system of equations
___
By Cramer's rule:
x = (3(11.75) -2(13.25))/(3(5) -2(4)) = 8.75/7 = 1.25
y = (13.25(5) -11.75(4))/7 = 19.25/7 = 2.75
The cost of one cookie is $1.25; the cost of one cupcake is $2.75.
_____
Cramer's rule gives the solution to the system of equations ...
ax +by =c
dx +ey = f
as ...
∆ = bd -ea
x = (bf -ec)/∆
y = (cd -fa)/∆
You might be interested in
This appears to be about rules of exponents as much as anything. The applicable "definitions, identities, and properties" are
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i
i^0 = 1 . . . . . as is true for any non-zero value to the zero power
i^1 = i . . . . . . as is true for any value to the first power
i^2 = -1 . . . . . from the definition of i
i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4
1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81
2. = i^((3-5)·2+0 = i^-4 = 1
3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i
4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i