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Understanding algebra is essential because it serves as the basis for more complex mathematical concepts. Algebra involves finding unknown values by using known mathematical relationships. In our exercise, we begin with the equation \( \frac{36}{y^2} = 1 \), which is a type of rational equation.

• Rational equations are equations that involve fractions where the numerator and the denominator are polynomials.
• In the given problem, the expression \( \frac{36}{y^2} \) is rational because \( y^2 \)—a polynomial—is in the denominator.

To solve the equation, the goal is to remove the fraction by isolating terms that include the variable \( y \). By multiplying both sides of the equation by \( y^2 \), algebra lets us simplify and clear the fraction, leading us closer to finding the solution. This process demonstrates how algebra simplifies expressions to make solving equations more straightforward.

Solving equations involves finding the value of the variable that makes the equation true. In algebra, we start with a given equation, and our journey toward finding the solutions requires careful manipulation of mathematical expressions. In this context:- The first step involves eliminating the fraction. We do this by multiplying both sides by the denominator \( y^2 \), which simplifies the equation to \( 36 = y^2 \).- Once the equation is in this simplified form, you'll notice it transforms into a much simpler problem, known as a quadratic equation.A quadratic equation is an equation where the highest exponent of the variable is 2. The equation \( y^2 = 36 \) is a perfect example, and solving it involves basic algebraic techniques. Solving such an equation involves taking the square root of both sides, allowing us to progress toward the final solution.

Square roots are a fundamental concept in algebra, helping solve equations where variables are squared. The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). This principle is applied in the exercise as follows:- Once you've simplified the original rational equation to \( y^2 = 36 \), finding the value of \( y \) involves calculating the square roots of both sides.- The square root of \( 36 \) is \( 6 \), but because \( (6)^2 = 36 \) and \( (-6)^2 = 36 \), both \( 6 \) and \( -6 \) are solutions.Understanding the concept of square roots allows one to solve equations efficiently by identifying multiple possible values of the variable. In this case, the presence of \( \pm \) indicates that both the positive and negative solutions are viable for this problem, reflecting a complete understanding of square roots.