Practice Internal - Systems of Equations

Phase 1: Le Restaurant

Le Restaurant prepares three set menus: Basic ($b$), Classic ($c$), and Deluxe ($d$). They use three resources daily: Pasta, Sauce, and Cheese.

Basic: 1 unit Pasta, 2 units Sauce, 3 units Cheese.
Classic: 2 units Pasta, 3 units Sauce, 5 units Cheese.
Deluxe: 3 units Pasta, 4 units Sauce, 6 units Cheese.

They have exactly 100 Pasta, 160 Sauce, and 250 Cheese available.

Form the equations and find the unique solution using your GC.

Pasta: $b$ + $c$ + $d$ =

Sauce: $b$ + $c$ + $d$ =

Cheese: $b$ + $c$ + $d$ =

b =

c =

d =

Incorrect. Check your columns and GC input. Attempt /3.

Solution

Eq 1: $1b + 2c + 3d = 100$
Eq 2: $2b + 3c + 4d = 160$
Eq 3: $3b + 5c + 6d = 250$
Solution: $b=30, c=20, d=10$

Achieved Passed!

Phase 2: The Shortage

A severe pasta shortage hits. The Chef alters the Pasta requirement for the menus to 2 for Basic, 4 for Classic, and 6 for Deluxe. They manage to secure exactly 250 Pasta.

Algebraic Proof

Original Pasta (Eq 1): $1b + 2c + 3d = 100$
NEW Pasta: $2b + 4c + 6d = 250$

Multiply original Eq 1 by 2, then subtract the NEW Pasta equation:

$0 =$

Incorrect. $200 - 250 = ?$ Attempt /3.

Solution

$200 - 250 = -50$

Geometry Visualization

Look at how the Original Pasta (Blue) and New Pasta (Red) equations interact in 3D space.

Fill in the blank:

"The system is inconsistent because the Original Pasta and New Pasta equations form planes, resulting in possible solution."

Incorrect. Check your terms. Attempt /3.

Solution

PARALLEL planes, resulting in NO (or ZERO) possible solution.

Merit Passed!

Phase 3: The Manager's Reset

The Manager steps in. Pasta is returned to normal. But now the Cheese recipe changes to 3 for Basic, 5 for Classic, and 7 for Deluxe. They order 260 Cheese.

(1) Pasta: $1b + 2c + 3d = 100$
(2) Sauce: $2b + 3c + 4d = 160$
(3) New Cheese: $3b + 5c + 7d = 260$

Prove Dependency

Add Eq(1) and Eq(2) together, then subtract Eq(3).

Eq(1) + Eq(2): $3b + 5c + 7d = 260$ Subtract Eq(3): $-(3b + 5c + 7d = 260)$

$=$

Incorrect. Attempt /3.

Solution

$0 = 0$

Form the General Solution

Let $d = d$. Take $2 \times Eq(1)$, then subtract $Eq(2)$ to isolate $c$.

$2(1b+2c+3d=100) \Rightarrow 2b+4c+6d=200$
$-(2b+3c+4d=160)$
-------------------------
$1c + 2d = 40$

$c = $ $d + 40$

Substitute $c$ into Eq(1): $1b + 2(40-2d) + 3d = 100 \Rightarrow b - d + 80 = 100$

$b = $ $d + 20$

Incorrect algebra. Attempt /3.

Solution

$c = -2d + 40$
$b = 1d + 20$ (or just $d+20$)

General Solution: $(d + 20,\; -2d + 40,\; d)$

Phase 3: Geometry & Reality

Use the 3D model below to verify the geometry of the dependent equations. Then use your General Solution to find a practical scenario for the chefs.

The Intersection Line

Identify the Geometry:

"The system is dependent because the 3 equation planes intersect along a single ."

Incorrect. Attempt /3.

Solution

They intersect along a LINE.

Specific Contextual Solution

Using your General Solution $(d + 20,\; -2d + 40,\; d)$, find a valid menu combination if the chefs decide to make exactly $15$ Deluxe meals ($d = 15$).

Classic ($c$) $= -2(15) + 40 =$

Basic ($b$) $= (15) + 20 =$

Incorrect arithmetic. Attempt /3.

Solution

$c = -30 + 40 = 10$
$b = 15 + 20 = 35$

Full Excellence Achieved!