Algebra - "Solve for x" Challenge

Problem Statement

Solve for \(x\) and \(y\). You should have coordinate pairs (\(x\), \(y\)) for each equation:

\[ x + y = 16 \]

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \]

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Step-by-Step Solution

To solve the system of equations:

\[ \text{equation 1: } x + y = 16 \]

\[ \text{equation 2: } \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \]

We will solve this step by step.

Step 1: Express \( y \) in terms of \( x \) from Equation 1

From Equation 1, we can isolate \( y \):

\[ x + y = 16 \]

\[ y = 16 - x \]

Step 2: Substitute \( y = 16 - x \) into Equation 2

Now, substitute \( y = 16 - x \) into Equation 2:

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \]

Substitute \( y = 16 - x \):

\[ \frac{1}{x} + \frac{1}{16 - x} = \frac{1}{3} \]

Step 3: Combine the fractions on the left-hand side

To combine the fractions \(\frac{1}{x} + \frac{1}{16 - x}\), find a common denominator:

\[ \frac{1}{x} + \frac{1}{16 - x} = \frac{(16 - x) + x}{x(16 - x)} = \frac{16}{x(16 - x)} \]

Now, the equation becomes:

\[ \frac{16}{x(16 - x)} = \frac{1}{3} \]

Step 4 Cross-multiply to eliminate the fractions

Cross-multiply to eliminate the fractions:

\[ 16 \times 3 = x(16 - x) \]

Simplifying:

\[ 48 = 16x - x^2 \]

Step 5: Rearrange the equation into standard quadratic form

Rearrange the equation:

\[ x^2 - 16x + 48 = 0 \]

Step 6: Solve the quadratic equation

Now, solve the quadratic equation \(x^2 - 16x + 48 = 0\). To do this, use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a = 1\), \(b = -16\), and \(c = 48\).

First, calculate the discriminant:

\[ \Delta = (-16)^2 - 4(1)(48) = 256 - 192 = 64 \]

Now, apply the quadratic formula:

\[ x = \frac{-(-16) \pm \sqrt{64}}{2(1)} = \frac{16 \pm 8}{2} \]

So, the two possible values for \(x\) are:

\[ x = \frac{16 + 8}{2} = \frac{24}{2} = 12 \]

or

\[ x = \frac{16 - 8}{2} = \frac{8}{2} = 4 \]

Step 7: Find corresponding values of \(y\)

For \(x = 12\), substitute into \(y = 16 - x\):

\[ y = 16 - 12 = 4 \]

For \(x = 4\), substitute into \(y = 16 - x\):

\[ y = 16 - 4 = 12 \]

Solution:

The solutions to the system are:

\[ (x, y) = (12, 4) \quad \text{or} \quad (x, y) = (4, 12) \]

Both solutions satisfy the system of equations.