Problem 139
Question
For problems \(57-140\), solve each equation. $$ \frac{-3 m}{2}+1=5 m $$
Step-by-Step Solution
Verified Answer
The solution is \( m = \frac{2}{13} \).
1Step 1: Eliminate Fractions by Multiplication
To eliminate the fraction, multiply both sides of the equation by 2, which is the denominator of the fraction. This gives:\[ 2 \left( \frac{-3m}{2} + 1 \right) = 2 \times 5m \]Simplifying gives:\[ -3m + 2 = 10m \]
2Step 2: Isolate Variable Terms on One Side
Add \(3m\) to both sides of the equation to get all \(m\) terms on one side:\[ -3m + 3m + 2 = 10m + 3m \]Simplifying gives:\[ 2 = 13m \]
3Step 3: Solve for the Variable
Divide both sides of the equation by 13 to solve for \(m\):\[ \frac{2}{13} = m \]
4Step 4: Write Final Answer
The solution for the equation is:\[ m = \frac{2}{13} \]
Key Concepts
Eliminating FractionsIsolating VariablesLinear EquationsVariable Terms
Eliminating Fractions
Fractions in equations can seem daunting, but they can be managed easily by eliminating them. One effective method is to multiply each term by the denominator of the fraction. This clears all the fractions and makes the equation simpler. For instance, consider the equation \( \frac{-3m}{2} + 1 = 5m \). The fraction \( \frac{-3m}{2} \) has a denominator of 2. Multiplying both sides of the equation by 2 will eliminate the fraction:
- Multiply every term by 2: \( 2 \times \left( \frac{-3m}{2} + 1 \right) = 2 \times 5m \)
- This results in: \( -3m + 2 = 10m \)
Isolating Variables
Once the fractions are eliminated, the next step is to isolate the variable on one side of the equation. This makes it easier to solve for the variable. The key is to perform the same operation on both sides to maintain the balance of the equation.In our example, the equation is now:\( -3m + 2 = 10m \). We want all the terms involving \( m \) on one side:
- Add \( 3m \) to both sides to compensate for the negative sign: \( -3m + 3m + 2 = 10m + 3m \)
- This simplifies to: \( 2 = 13m \)
Linear Equations
Linear equations are equations where the variable is raised to the power of one. They are expressed in the form \( Ax + B = C \), where \( A \), \( B \), and \( C \) are constants and \( x \) is the variable.A linear equation aims to find the specific value of the variable that makes the equation true. These are foundational equations in algebra, focusing on operations such as:
- Addition and subtraction of constants
- Multiplication and division by coefficients
- Rearranging terms to isolate the variable
Variable Terms
The term 'variable' refers to the symbol representing an unknown value in mathematical expressions and equations. In our example equation development, 'm' is the variable term.The main goal in equations with variables is to solve for the variable, finding its specific value. This involves:
- Combining like terms to simplify the equation
- Using operations such as addition, subtraction, multiplication, and division
- Ultimately isolating the variable
Other exercises in this chapter
Problem 137
For problems \(57-140\), solve each equation. $$ \frac{5 y}{13}-4=\frac{7 y}{26}+1 $$
View solution Problem 138
For problems \(57-140\), solve each equation. $$ \frac{-3 m}{5}=\frac{6 m}{10}-2 $$
View solution Problem 140
For problems \(57-140\), solve each equation. $$ -3 z=\frac{2 z}{5} $$
View solution Problem 136
For problems \(57-140\), solve each equation. $$ \frac{3 x}{7}=\frac{-3 x}{7}+12 $$
View solution