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Chapter 0: Problem 22

Solve the equation and check your solution. $$ \frac{x}{5}-\frac{x}{2}=3+\frac{3 x}{10} $$

Short Answer

Expert verified
The solution to the equation is \(x = -5\).

Step by step solution

01

Remove Fractions

Multiplying each term by 10 to remove the fractions, we get $2x - 5x = 30 + 3x$.
02

Consolidate like terms

The next step is to combine like terms, we get $-3x - 3x = 30$. So, it simplifies to $-6x = 30$.
03

Solve for x

Divide each side by -6, we get \(x = -\frac{30}{6}\). So, \(x = -5\).
04

Verify solution

Substitute x = -5 back into the original equation. Doing this we get \(-1 - (-5) = 3 - 1.5\). Both sides simplify to 2, so the solution is correct.

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Key Concepts

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

Solving Linear Equations
When you encounter a linear equation like the one in our exercise, the main goal is to find the value of the variable that makes the equation true. This process generally involves a few key steps such as clearing any fractions, combining like terms, isolating the variable, and simplifying the equation at each stage. It's like a puzzle where the rules of arithmetic and algebra guide you to the solution. Remember, what you do to one side of the equation, you do to the other to maintain balance.
Combining Like Terms
It's important to recognize terms that are 'like' or similar because they have the same variable raised to the same power. In our exercise, you learned that \(2x\) and \(5x\) are like terms, so they can be combined. This process simplifies the equation and gets you one step closer to isolating the variable. When combining like terms, you add or subtract the coefficients (the numbers in front of the variables) while keeping the variable part the same.
Verifying Solutions
After finding a potential solution to an equation, it is crucial to verify it by substituting the value back into the original equation. This step checks the work and ensures that the solution is correct. Our exercise demonstrated this with \(x = -5\) giving both sides of the equation an identical value when substituted, confirming the solution's validity. This step might seem redundant, but it is essential for avoiding mistakes.
Fractions in Equations
Equations containing fractions might seem more challenging, but they are just like any other equation once you clear the fractions. A common technique is to multiply every term by the least common denominator to remove the fractions, which was done in the first step of our exercise solution. This technique transforms a tricky fractional equation into a simpler one that's often easier to solve.

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Most popular questions from this chapter

True or False? In Exercises 85 and 86, determine whether the statement is true or false. Justify your answer. If the inverse function of \(f\) exists and the graph of \(f\) has a \(y\)-intercept, the \(y\)-intercept of \(f\) is an \(x\)-intercept of \(f^{-1}\).

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ p(x)=-4 $$

Determine whether the function is even, odd, or neither. Then describe the symmetry. $$ f(t)=t^{2}+2 t-3 $$

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)=\sqrt{x-2} $$

The function given by $$ y=0.03 x^{2}+245.50, \quad 0
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