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Chapter 3: Problem 3

Find the real solutions to the equation. $$ \frac{2 x+1}{x-5}=0 $$

Short Answer

Expert verified
The real solution is \(x = -\frac{1}{2}\).

Step by step solution

01

Set the Numerator Equal to Zero

For the fraction \(\frac{2x+1}{x-5} \) to be zero, the numerator must be zero. So, set the numerator equal to zero:\(2x + 1 = 0\).
02

Solve for x

Isolate x by subtracting 1 from both sides:\(2x + 1 - 1 = 0 - 1\), which simplifies to: \2x = -1\. Next, divide both sides by 2 to solve for x: \x = -\frac{1}{2}\.
03

Verify the Solution

Verify that the solution does not make the denominator zero. Substitute \x = -\frac{1}{2}\ into the denominator \(x - 5 \) to ensure it is not zero:\-\frac{1}{2} - 5 \eq 0\.Thus, the solution is valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Setting Numerator to Zero
To solve a rational equation like \(\frac{2 x+1}{x-5}=0\), you should start by setting the numerator equal to zero. This is because a fraction is only zero when its numerator is zero, and the denominator is not zero. That鈥檚 the key idea to remember for rational equations.
In our example, the numerator is \(2x + 1\). So, we set it to zero:
\2x + 1 = 0\.
Now, we鈥檒l solve this equation to find the value of \x\.
Isolating Variable
After setting the numerator to zero and having \( 2x + 1 = 0 \), our next step is to isolate the variable \( x \).
First, subtract 1 from both sides to get:
\2x + 1 - 1 = 0 - 1\.
This simplifies to:
\2x = -1\.
Next, divide both sides by 2:
\x = -\frac{1}{2}\.
Now, we have isolated \( x \) and found that \( x = -\frac{1}{2} \).

This means the proposed solution is \( x = -\frac{1}{2} \). However, we need to verify this solution.
Verifying Solution
The final, crucial step is to verify that the solution \( x = -\frac{1}{2} \) does not make the denominator zero. A rational equation is valid only if the denominator is not zero for the solution.
Let's substitute \( x = -\frac{1}{2} \) into the denominator \( x - 5 \):
\- \frac{1}{2} - 5 eq 0\.
This simplifies to \( -\frac{1}{2} - 5 = -\frac{1}{2} - \frac{10}{2} = -\frac{11}{2}\), which is clearly not zero.
Since the denominator is not zero, the solution \( x = -\frac{1}{2} \) is verified and valid.
  • First, ensure the numerator is zero by setting it to zero.
  • Isolate the variable by solving the equation.
  • Finally, check the solution by ensuring the denominator is not zero.
Remembering these steps will help you tackle and solve rational equations with confidence!

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Most popular questions from this chapter

The number of adults in U.S. prisons and jails for the years \(1980-2008\) is shown in the graph. (Source: U.S. Department of Justice, www.justice.gov) The variable \(t\) represents the number of years since 1980 . The function defined by \(P(t)=-0.091 t^{3}+3.48 t^{2}+15.4 t+335\) represents the number of adults in prison \(P(t)\) (in thousands). The function defined by \(J(t)=23.0 t+159\) represents the number of adults in jail \(J(t)\) (in thousands). a. Write the function defined by \(N(t)=(P+J)(t)\) and interpret its meaning in context. b. Write the function defined by \(R(t)=\left(\frac{J}{N}\right)(t)\) and interpret its meaning in the context of this problem. c. Evaluate \(R(25)\) and interpret its meaning in context. Round to 3 decimal places.

The number of U.S. citizens of voting age, \(N(t)\) (in millions), can be modeled according to the number of years \(t\) since 1932 . $$N(t)=0.0135 t^{2}+1.09 t+73.2$$ The function defined by $$V(t)=1.08 t+36.9$$ represents the number of people who voted \(V(t)\) (in millions) in U.S. presidential elections ( \(t\) is the number of years since 1932 and \(t\) is a multiple of 4). (Source: U.S. Census Bureau, www.census.gov) a. Write the function defined by \(P(t)=\left(\frac{V}{N}\right)(t)\) and interpret its meaning in the context of this problem. b. Evaluate \(P(60)\) and interpret its meaning in the context of this problem. Round to 2 decimal places.

Determine if the statement is true or false. If 5 is an upper bound for the real zeros of \(f(x)\), then 4 is also an upper bound.

Explain why \(x=-2\) is not a vertical asymptote of the graph of \(f(x)=\frac{x^{2}+7 x+10}{x+2}\).

Write the domain of the function in interval notation. $$ g(t)=\sqrt{1-t^{2}} $$
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