et ‘x’ and ‘y’ be real numbers such that |x – y| + 3x – y = 70, and |y – x| + 3y + x = 50. Then, the possible value of (x + 2y)/5

1. Break Down the Problem

We have two equations involving absolute values:

1. $|x - y| + 3x - y = 70$
2. $|y - x| + 3y + x = 50$

From the properties of absolute values, note that $|x - y| = |y - x|$. We can analyze two cases for the first equation based on the sign of $(x - y)$.

2. Relevant Concepts

We will consider two cases for $|x - y|$:

• Case 1: $x - y \geq 0$ ⇒ $|x - y| = x - y$
• Case 2: $x - y < 0$ ⇒ $|x - y| = y - x$

3. Analysis and Detail

Case 1: $x - y \geq 0$

• Equation 1 becomes: $(x - y) + 3x - y = 70 \implies 4x - 2y = 70 \implies 2x - y = 35 \quad \text{(1)}$

• Equation 2 becomes: $(y - x) + 3y + x = 50 \implies -x + 4y = 50 \implies x - 4y = -50 \quad \text{(2)}$

Solve Equations (1) and (2):

From (1): y = 2x - 35 \tag{3}

Substituting (3) into (2): $x - 4(2x - 35) = -50 \implies x - 8x + 140 = -50 \implies -7x + 140 = -50 \implies -7x = -50 - 140 \implies -7x = -190 \implies x = \frac{190}{7} \approx 27.14$ Substituting $x$ back into (3): $y = 2\left(\frac{190}{7}\right) - 35 = \frac{380}{7} - \frac{245}{7} = \frac{135}{7} \approx 19.29$

Now, we can calculate $\frac{x + 2y}{5}$: $\frac{x + 2y}{5} = \frac{\left(\frac{190}{7} + 2 \cdot \frac{135}{7}\right)}{5} = \frac{\left(\frac{190 + 270}{7}\right)}{5} = \frac{\left(\frac{460}{7}\right)}{5} = \frac{460}{35} = \frac{92}{7} \approx 13.14$

Case 2: $x - y < 0$

This implies $y > x$, so we have:

• Equation 1: $(y - x) + 3x - y = 70 \implies -x + 2x = 70 \implies x = 70 \quad \text{(4)}$

• Equation 2: $(x - y) + 3y + x = 50 \implies -y + 4y = 50 \implies y = 50 \quad \text{(5)}$

Using (4) and (5), we check the values:

• $x = 70$
• $y = 50$

Calculate $\frac{x + 2y}{5}$: $\frac{x + 2y}{5} = \frac{70 + 2 \cdot 50}{5} = \frac{70 + 100}{5} = \frac{170}{5} = 34$

4. Verify and Summarize

From both cases, we have:

• Case 1 gives $\frac{92}{7}$
• Case 2 gives $34$

Final Answer

The possible values of $\frac{x + 2y}{5}$ are: $\frac{92}{7} \quad \text{and} \quad 34$