4. $$ \left\{\begin{array}{l}x-10 y=60 \ x+14 y=12\end{array} ight. $$ 5. $$ \left\{\begin{array}{l}5 x+y=0 \ 5 x+2 y=30\end{array} ight. $$ 6. $$ \left\{\begin{array}{l}-5 x+7 y=11 \ -5 x+3 y=19\end{array} ight. $$ 8. $$ \left\{\begin{array}{l}-3 x+4 y=12 \ 2 x+y=-8\end{array} ight. $$ 9. $$ \left\{\begin{array}{l}2 x+4 y=-4 \ 3 x+5 y=-3\end{array} ight. $$ 10. Consumer Economics Each family in a neighborhood is contributing $$ \$ 20 $$ of food to the neighborhood picnic. The Harlin family is bringing 12 packages buns. The hamburger buns cost $$ \$ 2.00 $$ per package. The hot-dog buns cost $$ \$ 1 $$, package. How many packages of each type of bun did they buy?

Question

4. $$ \left\{\begin{array}{l}x-10 y=60 \ x+14 y=12\end{array}ight. $$ 5. $$ \left\{\begin{array}{l}5 x+y=0 \ 5 x+2 y=30\end{array}ight. $$ 6. $$ \left\{\begin{array}{l}-5 x+7 y=11 \ -5 x+3 y=19\end{array}ight. $$ 8. $$ \left\{\begin{array}{l}-3 x+4 y=12 \ 2 x+y=-8\end{array}ight. $$ 9. $$ \left\{\begin{array}{l}2 x+4 y=-4 \ 3 x+5 y=-3\end{array}ight. $$ 10. Consumer Economics Each family in a neighborhood is contributing $$ \$ 20 $$ of food to the neighborhood picnic. The Harlin family is bringing 12 packages buns. The hamburger buns cost $$ \$ 2.00 $$ per package. The hot-dog buns cost $$ \$ 1 $$, package. How many packages of each type of bun did they buy?

UpStudy AI · Accepted Answer

Sure, let's solve each of the provided problems step by step.

---

### **Problem 4**
$$
\begin{cases}
x - 10y = 60 \
x + 14y = 12
\end{cases}
$$

**Solution:**

1. **Subtract the first equation from the second to eliminate $$ x $$:**
   $$
   (x + 14y) - (x - 10y) = 12 - 60 \
   x + 14y - x + 10y = -48 \
   24y = -48 \
   y = \frac{-48}{24} = -2
   $$

2. **Substitute $$ y = -2 $$ into the first equation:**
   $$
   x - 10(-2) = 60 \
   x + 20 = 60 \
   x = 60 - 20 \
   x = 40
   $$

**Answer:** $$ x = 40 $$, $$ y = -2 $$

---

### **Problem 5**
$$
\begin{cases}
5x + y = 0 \
5x + 2y = 30
\end{cases}
$$

**Solution:**

1. **Subtract the first equation from the second to eliminate $$ x $$:**
   $$
   (5x + 2y) - (5x + y) = 30 - 0 \
   5x + 2y - 5x - y = 30 \
   y = 30
   $$

2. **Substitute $$ y = 30 $$ into the first equation:**
   $$
   5x + 30 = 0 \
   5x = -30 \
   x = \frac{-30}{5} \
   x = -6
   $$

**Answer:** $$ x = -6 $$, $$ y = 30 $$

---

### **Problem 6**
$$
\begin{cases}
-5x + 7y = 11 \
-5x + 3y = 19
\end{cases}
$$

**Solution:**

1. **Subtract the first equation from the second to eliminate $$ x $$:**
   $$
   (-5x + 3y) - (-5x + 7y) = 19 - 11 \
   -5x + 3y + 5x - 7y = 8 \
   -4y = 8 \
   y = \frac{8}{-4} = -2
   $$

2. **Substitute $$ y = -2 $$ into the first equation:**
   $$
   -5x + 7(-2) = 11 \
   -5x - 14 = 11 \
   -5x = 25 \
   x = \frac{25}{-5} \
   x = -5
   $$

**Answer:** $$ x = -5 $$, $$ y = -2 $$

---

### **Problem 8**
$$
\begin{cases}
-3x + 4y = 12 \
2x + y = -8
\end{cases}
$$

**Solution:**

1. **Solve the second equation for $$ y $$:**
   $$
   2x + y = -8 \
   y = -8 - 2x
   $$

2. **Substitute $$ y = -8 - 2x $$ into the first equation:**
   $$
   -3x + 4(-8 - 2x) = 12 \
   -3x - 32 - 8x = 12 \
   -11x - 32 = 12 \
   -11x = 44 \
   x = \frac{44}{-11} = -4
   $$

3. **Find $$ y $$ using $$ x = -4 $$:**
   $$
   y = -8 - 2(-4) = -8 + 8 = 0
   $$

**Answer:** $$ x = -4 $$, $$ y = 0 $$

---

### **Problem 9**
$$
\begin{cases}
2x + 4y = -4 \
3x + 5y = -3
\end{cases}
$$

**Solution:**

1. **Multiply the first equation by 3 and the second by 2 to make the coefficients of $$ x $$ equal:**
   $$
   3(2x + 4y) = 3(-4) \Rightarrow 6x + 12y = -12 \
   2(3x + 5y) = 2(-3) \Rightarrow 6x + 10y = -6
   $$

2. **Subtract the second new equation from the first to eliminate $$ x $$:**
   $$
   (6x + 12y) - (6x + 10y) = -12 - (-6) \
   0x + 2y = -6 \
   y = \frac{-6}{2} = -3
   $$

3. **Substitute $$ y = -3 $$ into the first original equation:**
   $$
   2x + 4(-3) = -4 \
   2x - 12 = -4 \
   2x = 8 \
   x = \frac{8}{2} = 4
   $$

**Answer:** $$ x = 4 $$, $$ y = -3 $$

---

### **Problem 10: Consumer Economics**

**Problem Statement:**
Each family in a neighborhood is contributing \$20 worth of food to the neighborhood picnic. The Harlin family is bringing 12 packages of buns. The hamburger buns cost \$2.00 per package, and the hot-dog buns cost \$1.00 per package. How many packages of each type of bun did they buy?

**Solution:**

Let:
- $$ h $$ = number of hamburger bun packages
- $$ d $$ = number of hot-dog bun packages

**Given:**
1. Total packages of buns:
   $$
   h + d = 12 \quad \text{(Equation 1)}
   $$
2. Total cost of buns is \$20:
   $$
   2h + 1d = 20 \quad \text{(Equation 2)}
   $$

**Steps:**

1. **Express $$ d $$ from Equation 1:**
   $$
   d = 12 - h
   $$

2. **Substitute $$ d = 12 - h $$ into Equation 2:**
   $$
   2h + (12 - h) = 20 \
   2h + 12 - h = 20 \
   h + 12 = 20 \
   h = 20 - 12 \
   h = 8
   $$

3. **Find $$ d $$ using $$ h = 8 $$:**
   $$
   d = 12 - 8 = 4
   $$

**Answer:**
- **Hamburger bun packages:** 8
- **Hot-dog bun packages:** 4

---

If you have any further questions or need additional explanations, feel free to ask!
**Problem 4:** - $$ x = 40 $$ - $$ y = -2 $$ **Problem 5:** - $$ x = -6 $$ - $$ y = 30 $$ **Problem 6:** - $$ x = -5 $$ - $$ y = -2 $$ **Problem 8:** - $$ x = -4 $$ - $$ y = 0 $$ **Problem 9:** - $$ x = 4 $$ - $$ y = -3 $$ **Problem 10:** - Hamburger buns: 8 packages - Hot-dog buns: 4 packages

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Algebra Questions