Writing Linear Equations in Standard Form from Word Problems ~ Scaffolded - Free Printable

It looks like you've uploaded a worksheet titled "Writing Equations in Standard Form from Word Problems (Part 2)", which contains three word problems that require setting up equations in standard form (Ax + By = C) and solving them.

Since I can't view images directly, I’ll solve each problem based on the text you’ve provided. Let me go step by step and explain how to approach each one.

---

Problem 1:

> Your family is buying mountain salad and potato salad for the family reunion. The mountain salad costs $5.50 per pound and the potato salad costs $3.00 per pound. Write an equation that represents the cost of the salad your family buys for the reunion.

#### Step 1: Define variables
Let:
- $ x $ = number of pounds of mountain salad
- $ y $ = number of pounds of potato salad

#### Step 2: Write the cost equation
- Cost of mountain salad: $ 5.50x $
- Cost of potato salad: $ 3.00y $
- Total cost: $ 5.50x + 3.00y $

So, the equation representing the total cost is:

$$
5.50x + 3.00y = C
$$

Where $ C $ is the total cost. But since no specific total is given, this is the general equation.

Now, the next part asks:

> What are possible amounts of each salad that your family can buy for $127?

We now set $ C = 127 $:

$$
5.50x + 3.00y = 127
$$

We want possible whole-number solutions (since you can’t buy a fraction of a pound in practice).

Let’s solve for integer values.

Multiply both sides by 2 to eliminate decimals:

$$
11x + 6y = 254
$$

Now solve for integer solutions.

Try values of $ x $ and see if $ y $ is an integer.

Try $ x = 2 $:
$ 11(2) + 6y = 254 \Rightarrow 22 + 6y = 254 \Rightarrow 6y = 232 \Rightarrow y = 38.67 $ → not integer

Try $ x = 4 $:
$ 44 + 6y = 254 \Rightarrow 6y = 210 \Rightarrow y = 35 $ ✔

So one solution: $ x = 4 $, $ y = 35 $ → 4 lbs mountain, 35 lbs potato

Try $ x = 10 $:
$ 110 + 6y = 254 \Rightarrow 6y = 144 \Rightarrow y = 24 $ ✔

Try $ x = 16 $:
$ 176 + 6y = 254 \Rightarrow 6y = 78 \Rightarrow y = 13 $ ✔

Try $ x = 22 $:
$ 242 + 6y = 254 \Rightarrow 6y = 12 \Rightarrow y = 2 $ ✔

Try $ x = 28 $:
$ 308 > 254 $ → too big

So valid combinations:
- (4, 35)
- (10, 24)
- (16, 13)
- (22, 2)

You could also check $ x = 0 $: $ 6y = 254 \Rightarrow y = 42.33 $ → no
$ x = 6 $: $ 66 + 6y = 254 \Rightarrow 6y = 188 \Rightarrow y = 31.33 $ → no

So possible combinations are those listed above.

✔ Answer: Possible combinations include:
- 4 lbs mountain, 35 lbs potato
- 10 lbs mountain, 24 lbs potato
- 16 lbs mountain, 13 lbs potato
- 22 lbs mountain, 2 lbs potato

All satisfy $ 5.50x + 3.00y = 127 $

---

Problem 2:

> Your order is planning a wedding reception. She has been round tables that seat 8 people and rectangular tables that seat 10 people. She plans to have 200 people at the reception altogether. Write an equation that represents the possible tables that may be used to seat 200 people at the wedding.

Let:
- $ x $ = number of round tables (8 seats)
- $ y $ = number of rectangular tables (10 seats)

Total seating:
$$
8x + 10y = 200
$$

Simplify by dividing by 2:
$$
4x + 5y = 100
$$

Now find possible combinations of $ x $ and $ y $ such that both are non-negative integers.

Solve for $ x $:
$$
4x = 100 - 5y \Rightarrow x = \frac{100 - 5y}{4}
$$

We need $ 100 - 5y $ divisible by 4.

Try values of $ y $:

- $ y = 0 $: $ x = 100/4 = 25 $ ✔
- $ y = 4 $: $ 100 - 20 = 80 $, $ x = 20 $ ✔
- $ y = 8 $: $ 100 - 40 = 60 $, $ x = 15 $ ✔
- $ y = 12 $: $ 100 - 60 = 40 $, $ x = 10 $ ✔
- $ y = 16 $: $ 100 - 80 = 20 $, $ x = 5 $ ✔
- $ y = 20 $: $ 100 - 100 = 0 $, $ x = 0 $ ✔

Other values? Try $ y = 2 $: $ 100 - 10 = 90 $, $ 90/4 = 22.5 $ ✘
$ y = 6 $: $ 100 - 30 = 70 $, $ 70/4 = 17.5 $ ✘
Only even multiples of 4 work.

So valid pairs:
- (25, 0): 25 round, 0 rectangular
- (20, 4)
- (15, 8)
- (10, 12)
- (5, 16)
- (0, 20)

✔ Answer: Possible combinations:
- 25 round, 0 rectangular
- 20 round, 4 rectangular
- 15 round, 8 rectangular
- 10 round, 12 rectangular
- 5 round, 16 rectangular
- 0 round, 20 rectangular

All satisfy $ 8x + 10y = 200 $

---

Problem 3:

> You made gift baskets to sell at the craft show. Small baskets sell for $12, and large baskets sell for $22. Your goal is to sell $150 worth of baskets at the craft show. Write an equation that represents the amount of the baskets you have to sell to reach your goal.

Let:
- $ x $ = number of small baskets ($12 each)
- $ y $ = number of large baskets ($22 each)

Total revenue:
$$
12x + 22y = 150
$$

Simplify by dividing by 2:
$$
6x + 11y = 75
$$

Now find non-negative integer solutions.

Solve for $ x $:
$$
6x = 75 - 11y \Rightarrow x = \frac{75 - 11y}{6}
$$

Try values of $ y $:

- $ y = 0 $: $ x = 75/6 = 12.5 $ ✘
- $ y = 1 $: $ 75 - 11 = 64 $, $ 64/6 ≈ 10.67 $ ✘
- $ y = 2 $: $ 75 - 22 = 53 $, $ 53/6 ≈ 8.83 $ ✘
- $ y = 3 $: $ 75 - 33 = 42 $, $ 42/6 = 7 $ ✔ → $ x = 7 $
- $ y = 4 $: $ 75 - 44 = 31 $, $ 31/6 ≈ 5.17 $ ✘
- $ y = 5 $: $ 75 - 55 = 20 $, $ 20/6 ≈ 3.33 $ ✘
- $ y = 6 $: $ 75 - 66 = 9 $, $ 9/6 = 1.5 $ ✘
- $ y = 7 $: $ 75 - 77 = -2 $ → negative, stop

Only valid solution: $ y = 3 $, $ x = 7 $

Check: $ 12(7) + 22(3) = 84 + 66 = 150 $ ✔

Are there others?

Try $ y = 9 $: $ 11×9 = 99 > 75 $ → too big

So only one combination: 7 small, 3 large

But wait — try $ y = 3 $, $ x = 7 $ → yes

Is there another?

Wait — let's check $ y = 0 $ to $ y = 6 $ again.

What about $ y = 3 $: works

Any other?

Try $ y = 9 $? Too big.

Wait — what if we try $ y = 3 $, $ x = 7 $ → yes

Can we try $ y = 3 + 6k $? Since 6 and 11 are coprime, but let’s just test all.

Alternatively, use Diophantine method.

We have $ 6x + 11y = 75 $

We found $ x = 7 $, $ y = 3 $ is a solution.

General solution:
Since GCD(6,11)=1, and 1 divides 75, solutions exist.

The general solution is:
$$
x = 7 + 11t \\
y = 3 - 6t
$$

For integer $ t $

We need $ x \geq 0 $, $ y \geq 0 $

Try $ t = 0 $: $ x = 7 $, $ y = 3 $ ✔
$ t = 1 $: $ x = 18 $, $ y = -3 $ ✘
$ t = -1 $: $ x = -4 $, $ y = 9 $ ✘

So only solution is $ x = 7 $, $ y = 3 $

✔ Answer: Only possible combination is:
- 7 small baskets and 3 large baskets

Check: $ 7×12 = 84 $, $ 3×22 = 66 $, $ 84+66=150 $ ✔

---

✔ Final Answers Summary:

#### Problem 1:
- Equation: $ 5.50x + 3.00y = 127 $
- Possible combinations:
- (4, 35), (10, 24), (16, 13), (22, 2)

#### Problem 2:
- Equation: $ 8x + 10y = 200 $ or simplified $ 4x + 5y = 100 $
- Possible combinations:
- (25,0), (20,4), (15,8), (10,12), (5,16), (0,20)

#### Problem 3:
- Equation: $ 12x + 22y = 150 $ or $ 6x + 11y = 75 $
- Only solution: $ x = 7 $, $ y = 3 $
- So: 7 small baskets, 3 large baskets

---

Let me know if you'd like these formatted as a printable answer sheet or with graphs!