Problem 350 - Elementary Algebra

Question

Solve. \((3 y-5)(y+7)=0\)

Step-by-Step Solution

Answer

The solutions are \( y = \frac{5}{3} \) and \( y = -7 \).

Step 1: Understand the Zero-Product Property

The Zero-Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. Here, we have the equation \((3 y-5)(y+7)=0\). We must set each factor equal to zero and solve for \(y\).

Step 2: Set the First Factor to Zero

Set the first factor \((3 y-5)\) to zero and solve for \(y\): \[ 3y - 5 = 0 \] Add 5 to both sides: \[ 3y = 5 \] Divide both sides by 3: \[ y = \frac{5}{3} \]

Step 3: Set the Second Factor to Zero

Set the second factor \((y+7)\) to zero and solve for \(y\): \[ y + 7 = 0 \] Subtract 7 from both sides: \[ y = -7 \]

Step 4: Combine the Solutions

The solutions to the equation \((3 y-5)(y+7)=0\) are the values for \(y\) that we found in the previous steps. Thus, the solutions are: \[ y = \frac{5}{3} \] and \[ y = -7 \].

Key Concepts

solving quadratic equations

Quadratic equations are polynomial equations of degree two. The general form is: \[ ax^2 + bx + c = 0 \]Here, 'a', 'b', and 'c' are constants, and 'x' represents an unknown variable. Solving quadratic equations can be done by various methods including factoring, completing the square, and using the quadratic formula. In our case, the equation \((3 y-5)(y+7)=0\)is already factored, which allows us to use the Zero-Product Property to find the solutions.

factoring

Factoring is the process of breaking down a polynomial into simpler components, called factors, that when multiplied together give the original polynomial. For quadratic equations, factoring involves finding two binomials whose product matches the original quadratic equation. In the exercise \((3 y-5)(y+7)=0\), the quadratic equation is already factored. This means it has been broken down into two binomial factors: \(3 y-5\) and \(y+7\). By setting each factor equal to zero, we can use the Zero-Product Property to solve for the variable.

algebraic solutions

Algebraic solutions refer to finding the values of the variable that satisfy the equation using algebraic techniques. To solve the given equation \( (3 y-5)(y+7)=0 \),we apply the Zero-Product Property, which states that if the product of two factors is zero, then at least one of the factors must be zero. First, set \( 3 y-5 = 0 \) and solve: \[ 3y - 5 = 0 \] Add 5 to both sides:\[ 3y = 5 \] Then, divide both sides by 3:\[ y = \frac{5}{3} \]Next, set \( y+7 = 0 \) and solve: \[ y + 7 = 0 \] Subtract 7 from both sides:\[ y = -7 \]These algebraic manipulations give us the solutions: \( y = \frac{5}{3} \) and \( y = -7 \).