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Chapter 13: Problem 30

In Problems \(30-33,\) solve for \(x\). $$ 5=\frac{3 x-5}{2 x+3} $$

Short Answer

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Question: Solve the given rational equation for x. $$ 5=\frac{3x-5}{2x+3} $$ Answer: The value of x that makes the given equation equal is \(x=-\frac{20}{7}\).

Step by step solution

01

Eliminate the denominator

The given equation is: $$ 5=\frac{3x-5}{2x+3} $$ The denominator in the given fraction is \((2x+3)\). To eliminate the denominator, multiply both sides by the least common multiple (LCM) of the denominators, which is \((2x+3)\). $$ (5)(2x+3) =\frac{3x-5}{2x+3}(2x+3) $$
02

Simplify the equation

Now, distribute and simplify each side of the equation: $$ 10x + 15 = 3x - 5 $$
03

Solve for x

Next, we will solve the linear equation to find the value of x. Subtract 3x from both sides: $$ 7x + 15 = -5 $$ Subtract 15 from both sides: $$ 7x = -20 $$ Finally, divide by 7: $$ x = -\frac{20}{7} $$ So the solution to the given equation is \(x=-\frac{20}{7}\).

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Key Concepts

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

Eliminating Denominators
When solving rational equations, one of the first tasks is to eliminate the fractions to simplify the process of finding a solution. Fractions can be intimidating, but they become easier to manage once you know how to remove the denominators.
To eliminate the denominator, we look for the least common multiple (LCM) of the denominators. In our example, the only denominator present is \(2x+3\). We multiply every term in the equation by this expression, effectively getting rid of the division:
  • This action gives us an equation without fractions, making it less complex to handle.
  • It avoids the need to deal with separate terms, allowing us to treat the equation like a standard algebraic expression.
By multiplying through by the denominator, the rational fraction becomes a straightforward linear equation. This allows us to focus on solving for \(x\) without the distraction of fractions.
Algebraic Manipulation
Algebraic manipulation involves changing the form of an equation to make it easier to solve. Once we have eliminated the denominator, the equation becomes ripe for manipulation. Let's explore the steps to simplify the equation from our example:
Distribution is the first move. Since we have the expression \((5)(2x+3) = 3x-5\), we distribute the 5 across \(2x+3\) which results in \(10x + 15 = 3x - 5\):
  • Distribute multiplication helps to eliminate parentheses, simplifying terms on each side of the equation.
  • This lays out all terms plainly, preparing for straightforward operations like addition and subtraction.
Once distribution is complete, further manipulation like combining like terms or isolating \(x\) helps in refining the equation to its simplest form. Keep tracking changes step by step to avoid mistakes.
Linear Equations
Linear equations are equations of the first degree, meaning they involve no exponents higher than one. They form straight lines when graphed and are typically written in the form \(ax + b = c\).
In our example, after eliminating the denominator and performing algebraic manipulation, we end up with a linear equation: \(7x + 15 = -5\). Solving for \(x\) is a process of isolating the variable:
  • Begin by eliminating or reducing the constants from \(x\)'s side of the equation. In this case, subtract 3x from both sides.
  • Next, subtract 15 from each side to further isolate terms containing \(x\).
  • After arranging terms, we get \(7x = -20\). By dividing both sides by 7, we solve for \(x\).
The solution \(x = -\frac{20}{7}\) results from these operations. Linear equations are solved by systematically isolating the variable, making them efficient once you understand the steps.

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Most popular questions from this chapter

Put each expression in Exercises \(1-4\) into the form \(a(x) / b(x)\) for polynomials \(a(x)\) and \(b(x)\). $$ \frac{3}{2 x+6}+5 $$

Put the rational expressions into quotient form, identify any horizontal or slant asymptotes, and sketch the graph. $$ \frac{6 x^{3}+3 x^{2}+17 x+8}{2 x^{2}+x+5} $$

Use the division algorithm to find the quotient \(q(x)\) and the remainder \(r(x)\) so that \(a(x)=\) \(q(x) b(x)+r(x)\). $$ \frac{2 x^{2}-3 x+11}{x^{2}+7} $$

Use the Remainder Theorem (page 427 ) to find the value of the constant \(r\) making the equations in identities. $$ \begin{array}{r} \frac{x^{7}-2 x^{3}+1}{x-2}=q(x)+\frac{r}{x-2} \\ \text { where } q(x)=x^{6}+2 x^{5}+4 x^{4}+8 x^{3}+14 x^{2}+28 x+56 . \end{array} $$

Divide common factors from the numerator and denominator in the rational expressions . Are the resulting expressions equivalent to the original expressions? $$ \frac{x^{3}+6 x^{2}+9 x}{x^{2}+10 x+16} $$
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