4) Jack is buying gift bags for his daughter’s 5th birthday party. His budget is between $20 and $30. He has a choice of $2, $3 and $4 bags. If he buys $4 bags he will have $3 remaining, if he buys $3 bags he will have $2 remaining, and if he buys $2 bags he will have $1 remaining. What is

his budget?

Let's solve this step-by-step:

Let's assume that Jack bought x $4 bags, y $3 bags, and z $2 bags.

According to the given information:
- If he buys $4 bags, he will have $3 remaining.
- If he buys $3 bags, he will have $2 remaining.
- If he buys $2 bags, he will have $1 remaining.

We can translate these statements into equations:

x * $4 + $3 = Budget (equation 1)
y * $3 + $2 = Budget (equation 2)
z * $2 + $1 = Budget (equation 3)

We also know that Jack's budget is between $20 and $30. So, we can add another equation to represent this:

Budget ≥ $20 (equation 4)
Budget ≤ $30 (equation 5)

Now, we can solve these equations to find the values of x, y, z, and the budget.

Substituting equation 1, 2, and 3 into equation 4 and 5, we get:

x * $4 + $3 ≥ $20 (equation 6)
y * $3 + $2 ≥ $20 (equation 7)
z * $2 + $1 ≥ $20 (equation 8)

x * $4 + $3 ≤ $30 (equation 9)
y * $3 + $2 ≤ $30 (equation 10)
z * $2 + $1 ≤ $30 (equation 11)

By solving these equations, we can determine the range of values for x, y, z, and the budget.

(Note: The specific values for x, y, z, and the budget will depend on the solutions obtained from solving these inequalities.)

I hope this step-by-step explanation helps!

To determine Jack's budget, we can use a bit of algebra. Let's start by assigning variables to the unknown values.

Let's call the number of $4 bags x, the number of $3 bags y, and the number of $2 bags z.

We are given three pieces of information:
1. If Jack buys $4 bags, he will have $3 remaining.
2. If Jack buys $3 bags, he will have $2 remaining.
3. If Jack buys $2 bags, he will have $1 remaining.

Using this information, we can set up three equations:

Equation 1: 4x + 3 = (his budget)
Equation 2: 3y + 2 = (his budget)
Equation 3: 2z + 1 = (his budget)

Now, we can set up a system of equations:

Equation 1: 4x + 3 = his budget ...(i)
Equation 2: 3y + 2 = his budget ...(ii)
Equation 3: 2z + 1 = his budget ...(iii)

We can simplify these equations by subtracting "his budget" from both sides:

Equation 1: 4x + 3 - his budget = 0 ...(iv)
Equation 2: 3y + 2 - his budget = 0 ...(v)
Equation 3: 2z + 1 - his budget = 0 ...(vi)

Now, let's solve this system of equations simultaneously. Subtracting equations (iv) and (v) gives:

4x + 3 - his budget - (3y + 2 - his budget) = 0
4x + 3 - 3y - 2 = 0
4x - 3y + 1 = 0 ...(vii)

Similarly, subtracting equations (v) and (vi) gives:

3y + 2 - his budget - (2z + 1 - his budget) = 0
3y + 2 - 2z - 1 = 0
3y - 2z + 1 = 0 ...(viii)

Now, let's subtract equations (vii) and (viii) to eliminate y:

(4x - 3y + 1) - (3y - 2z + 1) = 0
4x - 3y + 1 - 3y + 2z - 1 = 0
4x - 6y + z = 0 ...(ix)

Now, we will substitute the given information about the remaining amounts into equation (ix).
We know:
- If he buys $4 bags, he will have $3 remaining (equivalent to 4x - (his budget) + 3 = 0)
- If he buys $3 bags, he will have $2 remaining (equivalent to 6y - (his budget) + 2 = 0)
- If he buys $2 bags, he will have $1 remaining (equivalent to 4z - (his budget) + 1 = 0)

Substituting these values into equation (ix), we get:

4x - 6y + z = 0
4x - (4x + 3 - 3) - 6y + (6y + 2 - 2) + z - (4z + 1 - 1) = 0
4x - 4x - 3 + 3 - 6y + 6y + 2 - 2 + z - 4z - 1 + 1 = 0
0 = 0

We have obtained the equation 0 = 0, which means this equation is always true. This shows that the system of equations is dependent, and there are infinitely many solutions.

Therefore, we cannot determine Jack's budget from the given information.