Chapter 2: Problem 16
In \(3-20,\) solve each equation and check. $$ \frac{x}{2}=\frac{3}{2 x+1} $$
Short Answer
Expert verified
The solutions are \(x = \frac{3}{2}\) and \(x = -2\).
Step by step solution
01
Identify the Cross-Multiplication Need
The given equation is \( \frac{x}{2} = \frac{3}{2x + 1} \). This is a proportion, which can be solved by cross-multiplying to eliminate the fractions.
02
Cross Multiply
Cross-multiply the terms to get rid of the fractions: \[ x(2x + 1) = 3 \times 2 \] This simplifies to: \[ 2x^2 + x = 6 \]
03
Rearrange into a Quadratic Equation
Subtract 6 from both sides to form a standard quadratic equation: \[ 2x^2 + x - 6 = 0 \]
04
Factor the Quadratic Equation
We factor the quadratic equation \( 2x^2 + x - 6 = 0 \) into \[ (2x - 3)(x + 2) = 0 \] Here, we directly identify factors that satisfy the equation.
05
Solve for the Roots
Set each factor equal to zero and solve for \(x \): For \(2x - 3 = 0\), we get \(x = \frac{3}{2} \). For \(x + 2 = 0\), we get \(x = -2 \).
06
Check the Solutions
Substitute \(x = \frac{3}{2} \) into the original equation to check: \[ \frac{3/2}{2} = \frac{3}{4} \] and \[ \frac{3}{2 \times \frac{3}{2} + 1} = \frac{3}{4} \] Both sides are equal, so \(x = \frac{3}{2} \) is correct. Substitute \(x = -2\): \[ \frac{-2}{2} = -1 \] and \[ \frac{3}{2 \times -2 + 1} = 3/(-3) = -1 \] Both sides are equal, so \(x = -2\) is also correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross-Multiplication
Cross-multiplication is a powerful tool when dealing with equations that present a proportion. This technique allows us to eliminate fractions by multiplying across the equal sign. For example, in the equation \( \frac{x}{2} = \frac{3}{2x + 1} \), cross-multiplication involves multiplying the numerator of one side by the denominator of the other:
To apply this method, ensure that the equation forms a proportion and cross-multiply the terms to clear the fraction, producing a much simpler equation to manage.
- Multiply \( x \) by \( 2x + 1 \) on one side.
- Multiply \( 3 \) by \( 2 \) on the other side.
To apply this method, ensure that the equation forms a proportion and cross-multiply the terms to clear the fraction, producing a much simpler equation to manage.
Quadratic Equations
After applying cross-multiplication to our equation, we arrive at a quadratic equation: \( 2x^2 + x = 6 \). This form, characterized by an \( x^2 \) term, signifies we are dealing with a quadratic equation. Quadratic equations are polynomial equations of degree 2, and their standard form is \( ax^2 + bx + c = 0 \). In our case, the equation becomes \( 2x^2 + x - 6 = 0 \) after moving all terms to one side.
Understanding quadratic equations involves knowing their components:
Understanding quadratic equations involves knowing their components:
- \( x^2 \) is the quadratic term.
- \( x \) is the linear term.
- The constant term adjusts the equation to equal zero.
Factoring
Factoring is an important strategy for solving quadratic equations like \( 2x^2 + x - 6 = 0 \). When we factor, we're looking for two binomials whose product equals the quadratic equation. Here, the equation factors to \[ (2x - 3)(x + 2) = 0 \]Finding such factors often involves:
- Identifying two numbers that multiply to give the constant term \(-6\).
- Those numbers must also add to give the linear coefficient \(+1\).
- \( 2x - 3 = 0 \), yielding \( x = \frac{3}{2} \).
- \( x + 2 = 0 \), yielding \( x = -2 \).
Checking Solutions
After finding solutions for an equation, it's essential to check that they satisfy the original equation. For \( \frac{x}{2} = \frac{3}{2x + 1} \), checking each solution involves substituting back into the initial equation and ensuring both sides match.
- For \( x = \frac{3}{2} \), both fractions simplify to \( \frac{3}{4} \), confirming our solution is correct by both sides equating similarly.
- For \( x = -2 \), the left side simplifies to \(-1\) and the right to \(-1\), verifying this solution too, as both sides equate.