wo angies in an equilaterai triangie are $$ (a+b) $$ and $$ (2 a-10 b) $$ degrees nid the values of $$ a $$ and $$ b $$. ne sides of an equilateral triangle are $$ y-x) \mathrm{cm},(5 x+3) \mathrm{cm} $$ and $$ (2+2 y) \mathrm{cm} $$. nd the length of each side.

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### **Problem 1: Finding $$ a $$ and $$ b $$ in an Equilateral Triangle**

**Given:**
- Two angles in an equilateral triangle are $$ (a + b) $$ degrees and $$ (2a - 10b) $$ degrees.

**Understanding the Problem:**
- In an **equilateral triangle**, all three angles are equal, and each measures **60 degrees**.

**Setting Up Equations:**
1. Since both given expressions represent angles in the triangle:
   $$
   a + b = 60 \quad \text{(Equation 1)}
   $$
   $$
   2a - 10b = 60 \quad \text{(Equation 2)}
   $$

**Solving the System of Equations:**

1. **From Equation 1:**
   $$
   a = 60 - b
   $$
   
2. **Substitute $$ a $$ into Equation 2:**
   $$
   2(60 - b) - 10b = 60
   $$
   $$
   120 - 2b - 10b = 60
   $$
   $$
   120 - 12b = 60
   $$
   $$
   -12b = -60
   $$
   $$
   b = 5
   $$
   
3. **Find $$ a $$ using Equation 1:**
   $$
   a = 60 - 5 = 55
   $$

**Solution:**
$$
a = 55^\circ \quad \text{and} \quad b = 5^\circ
$$

---

### **Problem 2: Determining the Lengths of Sides in an Equilateral Triangle**

**Given:**
- The sides of an equilateral triangle are:
  $$
  y - x \ \text{cm}, \quad 5x + 3 \ \text{cm}, \quad \text{and} \quad 2 + 2y \ \text{cm}
  $$

**Understanding the Problem:**
- In an **equilateral triangle**, all three sides are equal in length.

**Setting Up Equations:**
1. Set the first side equal to the second:
   $$
   y - x = 5x + 3 \quad \text{(Equation 3)}
   $$
   
2. Set the first side equal to the third:
   $$
   y - x = 2 + 2y \quad \text{(Equation 4)}
   $$

**Solving the System of Equations:**

1. **From Equation 3:**
   $$
   y - x = 5x + 3
   $$
   $$
   y = 6x + 3 \quad \text{(Equation 5)}
   $$
   
2. **From Equation 4:**
   $$
   y - x = 2 + 2y
   $$
   $$
   -x - y = 2
   $$
   $$
   x + y = -2 \quad \text{(Equation 6)}
   $$
   
3. **Substitute Equation 5 into Equation 6:**
   $$
   x + (6x + 3) = -2
   $$
   $$
   7x + 3 = -2
   $$
   $$
   7x = -5
   $$
   $$
   x = -\frac{5}{7}
   $$
   
4. **Find $$ y $$ using Equation 5:**
   $$
   y = 6\left(-\frac{5}{7}\right) + 3 = -\frac{30}{7} + \frac{21}{7} = -\frac{9}{7}
   $$

**Calculating the Length of Each Side:**

Using $$ y - x $$:
$$
y - x = \left(-\frac{9}{7}\right) - \left(-\frac{5}{7}\right) = -\frac{4}{7} \ \text{cm}
$$

Since side lengths cannot be negative, we take the absolute value:
$$
\text{Length of each side} = \frac{4}{7} \ \text{cm}
$$

**Solution:**
$$
x = -\frac{5}{7}, \quad y = -\frac{9}{7}, \quad \text{and each side is} \ \frac{4}{7} \ \text{cm long}
$$

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### **Final Answers:**

1. **For the angles:**
   $$
   a = 55^\circ \quad \text{and} \quad b = 5^\circ
   $$

2. **For the sides of the equilateral triangle:**
   $$
   x = -\frac{5}{7}, \quad y = -\frac{9}{7}, \quad \text{and each side measures} \ \frac{4}{7} \ \text{cm}
   $$
$$ a = 55^\circ $$ and $$ b = 5^\circ $$. Each side of the equilateral triangle is $$ \frac{4}{7} $$ cm long.

wo angies in an equilaterai triangie are \( (a+b) \) and \( (2 a-10 b) \) degrees nid the values of \( a \) and \( b \). ne sides of an equilateral triangle are \( y-x) \mathrm{cm},(5 x+3) \mathrm{cm} \) and \( (2+2 y) \mathrm{cm} \). nd the length of each side.

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