3. $$ \frac{(-2 x y)^{5}(3 x y)^{2}\left(x^{-4}\right)}{(2 y)^{3}\left(-5 a^{2} b^{5}\right)^{0}}=\frac{\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}{\frac{1}{2}-\frac{1}{3}} $$ $$ M=(a+b)^{2}{ }^{2} N=\left(a^{2}+2 a b+b^{2}\right) \text { देतm: } a=-1 $$ $$ x-\frac{x}{7}=\frac{y}{8}+x+y=180 \quad \text { of } \frac{x}{8}=\frac{y}{3}+x-y $$

Question

UpStudy AI · Accepted Answer

Certainly! Let's tackle **Problem 3** step by step. It appears to contain multiple parts, so we'll address each one individually:

---

## **3(a)** Simplify the Expression

$$
\frac{(-2xy)^{5} \cdot (3xy)^{2} \cdot x^{-4}}{(2y)^{3} \cdot \left(-5a^{2}b^{5}\right)^{0}}
$$

### **Step 1: Simplify the Exponents**

1. **Expand each term with exponents:**

- $$(-2xy)^5 = (-2)^5 \cdot x^5 \cdot y^5 = -32x^5y^5$$
   - $$(3xy)^2 = 3^2 \cdot x^2 \cdot y^2 = 9x^2y^2$$
   - $$x^{-4} = \frac{1}{x^4}$$
   - $$(2y)^3 = 2^3 \cdot y^3 = 8y^3$$
   - $$\left(-5a^{2}b^{5}\right)^0 = 1$$ *(Any non-zero term raised to the power of 0 is 1)*

2. **Substitute the simplified terms back into the expression:**

$$
   \frac{(-32x^5y^5) \cdot (9x^2y^2) \cdot \frac{1}{x^4}}{8y^3 \cdot 1}
   $$

### **Step 2: Multiply the Numerators and Denominators**

1. **Numerator:**

$$
   (-32) \cdot 9 \cdot \frac{1}{x^4} = -288 \cdot \frac{1}{x^4}
   $$

And for the variables:

$$
   x^5 \cdot x^2 \cdot \frac{1}{x^4} = x^{5+2-4} = x^3
   $$
   
   $$
   y^5 \cdot y^2 = y^{5+2} = y^7
   $$

So, the numerator becomes:

$$
   -288x^3y^7
   $$

2. **Denominator:**

$$
   8y^3
   $$

### **Step 3: Combine Numerator and Denominator**

$$
\frac{-288x^3y^7}{8y^3} = -36x^3y^{7-3} = -36x^3y^4
$$

**Final Simplified Form:**

$$
-36x^{3}y^{4}
$$

---

## **3(b)** Evaluate Expressions Given $$ a = -1 $$

1. **Given Expressions:**

$$
   M = (a + b)^2 \quad \text{and} \quad N = a^2 + 2ab + b^2
   $$

2. **Given Values:**

$$
   a = -1
   $$

*(Note: The value of $$ b $$ isn't provided, so we'll express the results in terms of $$ b $$.)*

### **Calculate $$ M $$:**

$$
M = (a + b)^2
$$

Substitute $$ a = -1 $$:

$$
M = (-1 + b)^2 = (b - 1)^2 = b^2 - 2b + 1
$$

### **Calculate $$ N $$:**

$$
N = a^2 + 2ab + b^2
$$

Substitute $$ a = -1 $$:

$$
N = (-1)^2 + 2(-1)b + b^2 = 1 - 2b + b^2
$$

**Observation:**

Notice that:

$$
M = (b - 1)^2 = b^2 - 2b + 1 = N
$$

So, $$ M = N $$ when $$ a = -1 $$.

---

## **3(c)** Solve the System of Equations for $$ x $$ and $$ y $$

### **Given Equations:**

$$
\begin{cases}
x - \frac{x}{7} = 180 \
\frac{y}{8} + x + y = 180 \quad \text{and} \quad \frac{x}{8} = \frac{y}{3} + x - y
\end{cases}
$$

*However, the notation seems a bit unclear. Let's interpret it as two separate equations:*

1. **Equation 1:**

$$
   x - \frac{x}{7} = 180
   $$

2. **Equation 2:**

$$
   \frac{y}{8} + x + y = 180
   $$
   
3. **Additional Equation:**

$$
   \frac{x}{8} = \frac{y}{3} + x - y
   $$

*Assuming we have three equations, but typically, two variables require two equations. For clarity, we'll proceed with the first two equations.*

### **Equation 1: Solve for $$ x $$**

$$
x - \frac{x}{7} = 180
$$

Find a common denominator to combine like terms:

$$
\frac{7x}{7} - \frac{x}{7} = 180
$$

$$
\frac{6x}{7} = 180
$$

Multiply both sides by 7:

$$
6x = 180 \times 7 = 1260
$$

Divide both sides by 6:

$$
x = \frac{1260}{6} = 210
$$

### **Equation 2: Solve for $$ y $$**

$$
\frac{y}{8} + x + y = 180
$$

We know $$ x = 210 $$, substitute:

$$
\frac{y}{8} + 210 + y = 180
$$

Combine like terms:

$$
\frac{y}{8} + y = 180 - 210 = -30
$$

Convert $$ y $$ to eighths:

$$
\frac{y}{8} + \frac{8y}{8} = \frac{9y}{8} = -30
$$

Multiply both sides by 8:

$$
9y = -240
$$

Divide both sides by 9:

$$
y = -\frac{240}{9} = -\frac{80}{3} \approx -26.\overline{6}
$$

### **Summary of Solutions:**

$$
x = 210 \
y = -\frac{80}{3} \quad \text{or} \quad y \approx -26.\overline{6}
$$

---

**Note:** If the third equation $$\frac{x}{8} = \frac{y}{3} + x - y$$ is intended to be part of the system, additional steps are required to solve all three equations simultaneously. However, typically, two equations are sufficient for solving two variables ($$ x $$ and $$ y $$). If more clarity is provided on the equations, a more precise solution can be offered.
**Solutions:** 1. **Expression Simplification:** $$ -36x^{3}y^{4} $$ 2. **Given $$ a = -1 $$:** $$ M = N = b^2 - 2b + 1 $$ 3. **Solutions for $$ x $$ and $$ y $$:** $$ x = 210 \ y = -\frac{80}{3} \quad \text{or} \quad y \approx -26.\overline{6} $$

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