Problem 28
Question
Solve the equation and check your solution. (Some equations have no solution.) $$ \frac{x}{5}-\frac{x}{2}=3 $$
Step-by-Step Solution
Verified Answer
The solution to the given equation is \( x = -10 \).
1Step 1: Clear the Fractions
Firstly, to simplify the equation, remove the fractions. The LCM of the denominators \(2\) and \(5\) is \(10\). Therefore, multiplying every term by \(10\) gives: \(10 * (x/5) - 10 * (x/2) = 10 * 3\) which simplifies to \(2x - 5x = 30\).
2Step 2: Solve the Equation
Consolidate like terms on one side of the equals sign to get \( -3x = 30\). Divide each side of the equation by -3, the coefficient of \(x\), to isolate \(x\). Hence, \(x = 30 / -3\).
3Step 3: Verify the Solution
Check whether the obtained value of \( x \) when substituted back into the original equation satisfies the equation. For \( x = -10 \), the left side becomes \(-10/5 - (-10)/2 = -2+5=3\), which is indeed the right side.
4Step 4: Write the Final Solution
After verifying that the obtained solution works when substituted back into the original equation, it can be confirmed that \(x = -10\) is the solution to the equation.
Key Concepts
Solving Linear EquationsFractions in EquationsVerifying Solutions
Solving Linear Equations
Linear equations are equations where the variable is raised only to the first power. They usually have the form \(ax + b = c\). In the provided exercise, the equation is slightly more complex due to the presence of fractions, but it still represents a linear equation because the variable \(x\) appears only in the first degree.
To solve a linear equation, the goal is to isolate the variable on one side of the equation. This typically involves combining like terms and using inverse operations such as addition, subtraction, multiplication, or division. By following these steps, the complexity of handling different terms or coefficients can be simplified.
In our exercise, the steps to clear fractions and simplify the equation before isolating the variable are crucial to finding the solution efficiently.
To solve a linear equation, the goal is to isolate the variable on one side of the equation. This typically involves combining like terms and using inverse operations such as addition, subtraction, multiplication, or division. By following these steps, the complexity of handling different terms or coefficients can be simplified.
In our exercise, the steps to clear fractions and simplify the equation before isolating the variable are crucial to finding the solution efficiently.
Fractions in Equations
Fractions can make solving equations seem tricky, but they can be easily managed by using a strategic approach. Whenever you see fractions in an equation, a good step is to eliminate the fractions by finding a common denominator.
In the exercise, the denominators are \(5\) and \(2\), so the least common multiple (LCM) is \(10\). By multiplying every term in the equation by \(10\), we transform the equation into integer coefficients. This results in \(2x - 5x = 30\).
This elimination simplifies the work needed and brings the equation to a form that is easier to solve. It is essential to handle fractions diligently to ensure no errors in the transformation of the equation.
In the exercise, the denominators are \(5\) and \(2\), so the least common multiple (LCM) is \(10\). By multiplying every term in the equation by \(10\), we transform the equation into integer coefficients. This results in \(2x - 5x = 30\).
This elimination simplifies the work needed and brings the equation to a form that is easier to solve. It is essential to handle fractions diligently to ensure no errors in the transformation of the equation.
Verifying Solutions
After finding a potential solution for an equation, it is essential to verify that the solution satisfies the original problem. This step is crucial to ensure accuracy and correctness.
To verify the solution, substitute the solved value back into the original equation. If both sides of the equation balance or are equal, the solution is confirmed. For instance, in the exercise, substituting \(x = -10\) back into the original equation \(\frac{x}{5} - \frac{x}{2} = 3\), results in \( -2 + 5 = 3 \).
Verification is an important last step in solving any mathematical equation. It assures that no mistakes were made in calculations, ensuring the integrity of the solution.
To verify the solution, substitute the solved value back into the original equation. If both sides of the equation balance or are equal, the solution is confirmed. For instance, in the exercise, substituting \(x = -10\) back into the original equation \(\frac{x}{5} - \frac{x}{2} = 3\), results in \( -2 + 5 = 3 \).
Verification is an important last step in solving any mathematical equation. It assures that no mistakes were made in calculations, ensuring the integrity of the solution.
Other exercises in this chapter
Problem 28
Use the Quadratic Formula to solve the quadratic equation. $$ 25 h^{2}+80 h+61=0 $$
View solution Problem 28
Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ 9 x^{2}
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Solve the inequality. Then graph the solution set on the real number line. \(\frac{x+6}{x+1}
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Copy and complete the statement using the correct inequality symbol. If \(-\frac{2}{3} x \geq-6\), then \(x\)_______ \(9 .\)
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