Problem 47
Question
For the following problems, solve the rational equations. $$ \frac{12}{y}+\frac{12}{y^{2}}=-3 $$
Step-by-Step Solution
Verified Answer
Question: Solve the rational equation and find the value of y: $$\frac{12}{y} + \frac{12}{y^2} = -3$$
Answer: The solution to the given rational equation is \(y = -2\).
1Step 1: Find a common denominator
In this case, the common denominator for \(\frac{12}{y}\) and \(\frac{12}{y^2}\) is \(y^2\).
2Step 2: Combine the terms
Using the common denominator of \(y^2\), we can combine the terms like so:
$$
\frac{12y}{y^2} + \frac{12}{y^2} = -3
$$
Now, we can combine the numerators:
$$
\frac{12y + 12}{y^2} = -3
$$
3Step 3: Clear the denominator
To do this, multiply both sides of the equation by \(y^2\):
$$
(12y + 12) = -3y^2
$$
4Step 4: Rearrange and form a quadratic equation
Move all terms to the left side of the equation:
$$
3y^2 + 12y + 12 = 0
$$
5Step 5: Factor or use the quadratic formula
In this case, we can easily factor the quadratic equation:
$$
3(y^2 + 4y + 4) = 0
$$
$$
3(y + 2)^2 = 0
$$
Since it's factored, we can now solve for \(y\):
$$
(y + 2)^2 = 0
$$
6Step 6: Solve for \(y\)
In this case, there is only one solution for \(y\):
$$
y + 2 = 0 \Rightarrow y = -2
$$
So, the solution to the given rational equation is \(y = -2\).
Key Concepts
Quadratic EquationCommon DenominatorFactoringSolving Equations
Quadratic Equation
A quadratic equation is any equation that can be rearranged into the standard form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are encountered frequently in algebra and often describe parabolas in geometry. In this exercise, after clearing the fraction and simplifying, the equation \( 3y^2 + 12y + 12 = 0 \) is revealed to be quadratic.
Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{2a} \] Choosing the best method depends on the specific equation and sometimes on your personal preference. In this exercise, factoring is the suitable approach.
Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{2a} \] Choosing the best method depends on the specific equation and sometimes on your personal preference. In this exercise, factoring is the suitable approach.
Common Denominator
A common denominator is a common multiple of the denominators of two or more fractions. For rational equations, finding a common denominator allows you to combine fractions into a single expression.
In this particular exercise, we have two terms:
Using the common denominator \( y^2 \), we rewrite the equation to combine the fractions, easing the process of solving the rational equation. This step is crucial in order to deal with a single rational expression rather than separate terms.
In this particular exercise, we have two terms:
- \(\frac{12}{y}\)
- \(\frac{12}{y^2}\)
Using the common denominator \( y^2 \), we rewrite the equation to combine the fractions, easing the process of solving the rational equation. This step is crucial in order to deal with a single rational expression rather than separate terms.
Factoring
Factoring is a method used to break down expressions into simpler products of sums or differences. It’s particularly useful in simplifying polynomial equations, like quadratics.
In the exercise, the quadratic equation \(3(y^2 + 4y + 4) = 0\) is factored to find the solution. After dividing by 3, you recognize \((y + 2)^2\) as a perfect square trinomial.
Factoring the quadratic \(y^2 + 4y + 4\) gives:
Factoring simplifies solving, especially when equations are set to zero, because it reduces the problem to finding roots of simpler expressions.
In the exercise, the quadratic equation \(3(y^2 + 4y + 4) = 0\) is factored to find the solution. After dividing by 3, you recognize \((y + 2)^2\) as a perfect square trinomial.
Factoring the quadratic \(y^2 + 4y + 4\) gives:
- \((y + 2)(y + 2) = 0\)
- or simply, \((y + 2)^2 = 0\)
Factoring simplifies solving, especially when equations are set to zero, because it reduces the problem to finding roots of simpler expressions.
Solving Equations
Solving equations typically involves finding the value(s) of the variable(s) that satisfy the equation. In algebra, we encounter linear, quadratic, and rational equations among others.
This exercise involves solving a rational equation initially, which transforms into a quadratic equation. By factoring the quadratic to \((y + 2)^2 = 0\), you solve by finding the value of \(y\) that makes this expression zero.
The final step, solving \((y + 2)^2=0\), involves recognizing it has only one solution, \(y = -2\).
Consider these steps for solving equations:
This exercise involves solving a rational equation initially, which transforms into a quadratic equation. By factoring the quadratic to \((y + 2)^2 = 0\), you solve by finding the value of \(y\) that makes this expression zero.
The final step, solving \((y + 2)^2=0\), involves recognizing it has only one solution, \(y = -2\).
Consider these steps for solving equations:
- Clear any fractions by finding a common denominator
- Combine like terms
- Rearrange to isolate variable terms
- Factor (for quadratics) or use other suitable methods
- Calculate the roots or solutions
Other exercises in this chapter
Problem 47
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