Certainly! To solve the problem where you need to find pairs [tex]\((x, y)\)[/tex] that satisfy the equations:
1. [tex]\(\frac{1}{x} + \frac{1}{y} = \frac{7}{12}\)[/tex]
2. [tex]\(xy = 12\)[/tex]
we can follow these steps:
### Step 1: Simplify the equation [tex]\(\frac{1}{x} + \frac{1}{y} = \frac{7}{12}\)[/tex]
Using the equation [tex]\(\frac{1}{x} + \frac{1}{y} = \frac{7}{12}\)[/tex], we rewrite the left side with a common denominator:
[tex]\[
\frac{y + x}{xy} = \frac{7}{12}
\][/tex]
Substitute [tex]\(xy = 12\)[/tex] from the given:
[tex]\[
\frac{x + y}{12} = \frac{7}{12}
\][/tex]
### Step 2: Eliminate the denominators
To eliminate the denominators, multiply both sides of the equation by 12:
[tex]\[
x + y = 7
\][/tex]
### Step 3: Form a system of equations
Now, we have two equations:
[tex]\[
x + y = 7
\][/tex]
[tex]\[
xy = 12
\][/tex]
### Step 4: Solving the system of equations
The equations [tex]\(x + y = 7\)[/tex] and [tex]\(xy = 12\)[/tex] can be solved by treating [tex]\(x\)[/tex] and [tex]\(y\)[/tex] as roots of a quadratic equation. We consider a quadratic equation of the form:
[tex]\[
t^2 - (x + y)t + xy = 0
\][/tex]
Substitute [tex]\(x + y = 7\)[/tex] and [tex]\(xy = 12\)[/tex]:
[tex]\[
t^2 - 7t + 12 = 0
\][/tex]
### Step 5: Factor the quadratic equation
To solve for [tex]\(t\)[/tex], we factorize the quadratic equation:
[tex]\[
t^2 - 7t + 12 = 0
\][/tex]
This can be factored into:
[tex]\[
(t - 3)(t - 4) = 0
\][/tex]
### Step 6: Find the roots of the equation
Set each factor equal to zero and solve for [tex]\(t\)[/tex]:
[tex]\[
t - 3 = 0 \quad \text{or} \quad t - 4 = 0
\][/tex]
[tex]\[
t = 3 \quad \text{or} \quad t = 4
\][/tex]
### Step 7: Assign the solutions to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
Thus, the pairs [tex]\((x, y)\)[/tex] that satisfy both equations are:
[tex]\[
(x, y) = (3, 4) \quad \text{or} \quad (x, y) = (4, 3)
\][/tex]
So, the final solutions are:
[tex]\[
(x, y) = (3, 4) \quad \text{and} \quad (x, y) = (4, 3)
\][/tex]
