Question

find the solution to the system of equations
-3x + 14y = -28
-2x + 7y = -14

find the solution to the system of equations
x - y = -1
-10x + 14 = 2y
-4x + 4y = 24

the freshmen want to throw a party to may to celebrate surviving their first year. pizza hut charges a $5 delivery fee, plus $15 for each large pizza. alternatively, papa johns charges a $10 delivery fee, plus $12.50 for each large pizza. when will the cost of ordering from either company be the same? what is the total cost?

Explanation:

First System of Equations:

We have the system:

$$
LATEXBLOCK0
$$

Step 1: Multiply the second equation by 2

To eliminate \(x\) or \(y\), we can multiply the second equation by 2. The second equation is \(-2x + 7y = -14\). Multiplying both sides by 2 gives:
\(2\times(-2x + 7y)=2\times(-14)\)
\(-4x + 14y = -28\)

Now our system is:

$$
LATEXBLOCK1
$$

Step 2: Subtract the two equations

Subtract the first equation from the second equation (or vice versa) to eliminate \(y\). Let's subtract the first equation \(-3x + 14y = -28\) from the second equation \(-4x + 14y = -28\):

\((-4x + 14y)-(-3x + 14y)=-28-(-28)\)

Simplify the left side: \(-4x + 14y + 3x - 14y=-x\)

Simplify the right side: \(-28 + 28 = 0\)

So we have \(-x=0\), which means \(x = 0\)

Step 3: Substitute \(x = 0\) into one of the original equations

Let's substitute \(x = 0\) into the second original equation \(-2x + 7y=-14\):

\(-2(0)+7y=-14\)

Simplify: \(0 + 7y=-14\)

Divide both sides by 7: \(y=\frac{-14}{7}=-2\)

Second System of Equations:

We have the system:

$$
LATEXBLOCK2
$$

Step 1: Solve the first equation for \(x\)

From \(x - y=-1\), we can express \(x\) in terms of \(y\): \(x=y - 1\)

Step 2: Substitute \(x=y - 1\) into the second equation

Substitute \(x=y - 1\) into \(-10x + 14 = 2y\):

\(-10(y - 1)+14 = 2y\)

Step 3: Expand and simplify

Expand \(-10(y - 1)\): \(-10y+10 + 14=2y\)

Combine like terms: \(-10y+24 = 2y\)

Add \(10y\) to both sides: \(24=12y\)

Divide both sides by 12: \(y = 2\)

Step 4: Substitute \(y = 2\) into \(x=y - 1\)

\(x=2 - 1=1\)

Third Problem (Pizza Cost):

Let's define:

• Let \(n\) be the number of pizzas.
• Cost from Pizza Hut: \(C_{H}=5n + 15\) (a \$5 delivery fee plus \$15 for each large pizza)
• Cost from Papa John's: \(C_{J}=12.5n+0\) (no delivery fee, \$12.50 for each large pizza) Wait, wait, the text says "Pizza Hut charges a \$5 delivery fee, plus \$15 for each large pizza. Alternatively, Papa John's charges a \$12.50 delivery fee, plus \$7.50 for each large pizza". Oh, I misread. Let's correct:

Let \(n\) = number of pizzas.

Cost from Pizza Hut: \(C_{H}=5 + 15n\) (delivery fee \$5, \$15 per pizza)

Cost from Papa John's: \(C_{J}=12.5+7.5n\) (delivery fee \$12.5, \$7.5 per pizza)

We want to find when \(C_{H}=C_{J}\)

Step 1: Set the equations equal

\(5 + 15n=12.5+7.5n\)

Step 2: Subtract \(7.5n\) from both sides

\(5 + 7.5n=12.5\)

Step 3: Subtract 5 from both sides

\(7.5n=7.5\)

Step 4: Divide both sides by 7.5

\(n = 1\)

Now find the total cost by substituting \(n = 1\) into either \(C_{H}\) or \(C_{J}\)

Using \(C_{H}=5+15(1)=20\)

Using \(C_{J}=12.5 + 7.5(1)=20\)

Answer:

s:

1. For the first system: \(x = 0\), \(y=-2\)
2. For the second system: \(x = 1\), \(y = 2\)
3. For the pizza problem: The number of pizzas when the cost is the same is \(n = 1\), and the total cost is \$20.