`(1)/(2)r-(1)/(3)s=8`
`(5)/(8)r-(1)/(4)s=29`
For the system of equations above, what is the value of r+s?

Answer

To solve the system of equations given by:

1. \(\frac{1}{2}r - \frac{1}{3}s = 8\) (Equation 1)
2. \(\frac{5}{8}r - \frac{1}{4}s = 29\) (Equation 2)

we will follow these steps:

### Step 1: Eliminate the fractions in Equation 1
To eliminate the fractions in Equation 1, we can multiply the entire equation by 6 (the least common multiple of 2 and 3):

\[ 6 \left(\frac{1}{2}r\right) - 6 \left(\frac{1}{3}s\right) = 6 \cdot 8 \]

This simplifies to:

\[ 3r - 2s = 48 \quad \text{(Equation 3)} \]

### Step 2: Eliminate the fractions in Equation 2
Now, let's eliminate the fractions in Equation 2 by multiplying the entire equation by 8 (the least common multiple of 8 and 4):

\[ 8 \left(\frac{5}{8}r\right) - 8 \left(\frac{1}{4}s\right) = 8 \cdot 29 \]

This simplifies to:

\[ 5r - 2s = 232 \quad \text{(Equation 4)} \]

### Step 3: Subtract Equation 3 from Equation 4
Now, we will subtract Equation 3 from Equation 4 to eliminate \(s\):

\[ (5r - 2s) - (3r - 2s) = 232 - 48 \]

This simplifies to:

\[ 5r - 3r = 232 - 48 \]

\[ 2r = 184 \]

### Step 4: Solve for \(r\)
Now, divide both sides by 2 to find \(r\):

\[ r = \frac{184}{2} = 92 \]

### Step 5: Substitute \(r\) back into Equation 3 to find \(s\)
Now that we have \(r\), we can substitute \(r = 92\) back into Equation 3:

\[ 3(92) - 2s = 48 \]

This simplifies to:

\[ 276 - 2s = 48 \]

### Step 6: Solve for \(s\)
Now, isolate \(s\):

\[ -2s = 48 - 276 \]

\[ -2s = -228 \]

Dividing both sides by -2 gives:

\[ s = \frac{228}{2} = 114 \]

### Step 7: Find \(r + s\)
Finally, we can find \(r + s\):

\[ r + s = 92 + 114 = 206 \]

Thus, the value of \(r + s\) is **206**.

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