Math Problem Statement

Solve the equation: 3(2a/5 - 2) - 1 = 2(5 - 3a/4) + 1.

Solution

The equation in the image is as follows:

3(2a52)1=2(53a4)+13\left(\frac{2a}{5} - 2\right) - 1 = 2\left(5 - \frac{3a}{4}\right) + 1

Let’s solve step-by-step:


Step 1: Expand both sides

Expand the terms inside the parentheses:

Left-hand side: 3(2a52)=6a563\left(\frac{2a}{5} - 2\right) = \frac{6a}{5} - 6 So the left-hand side becomes: 6a561=6a57\frac{6a}{5} - 6 - 1 = \frac{6a}{5} - 7

Right-hand side: 2(53a4)=106a42\left(5 - \frac{3a}{4}\right) = 10 - \frac{6a}{4} Simplify: 103a210 - \frac{3a}{2} Add 1 to this: 103a2+1=113a210 - \frac{3a}{2} + 1 = 11 - \frac{3a}{2}


Step 2: Write the equation

The equation now is: 6a57=113a2\frac{6a}{5} - 7 = 11 - \frac{3a}{2}


Step 3: Eliminate fractions

The least common denominator (LCD) of 5 and 2 is 10. Multiply the entire equation by 10 to eliminate the fractions: 10(6a57)=10(113a2)10\left(\frac{6a}{5} - 7\right) = 10\left(11 - \frac{3a}{2}\right)

Simplify both sides: 12a70=11015a12a - 70 = 110 - 15a


Step 4: Combine like terms

Bring all terms involving aa to one side and constants to the other: 12a+15a=110+7012a + 15a = 110 + 70 27a=18027a = 180


Step 5: Solve for aa

Divide both sides by 27: a=18027=203a = \frac{180}{27} = \frac{20}{3}


Final Answer:

a=203a = \frac{20}{3}


Would you like a detailed explanation of any step?


5 Related Questions:

  1. How do you solve equations with fractions efficiently?
  2. What strategies are useful when expanding and simplifying terms in algebra?
  3. How can you determine the least common denominator of multiple fractions?
  4. Why is isolating the variable aa important in solving linear equations?
  5. How can you check if the solution a=203a = \frac{20}{3} is correct?

Tip:

When solving equations with fractions, always consider clearing the fractions by multiplying through by the least common denominator (LCD) to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Fractions

Formulas

Distributive property: a(b + c) = ab + ac
Clearing fractions using the least common denominator (LCD)

Theorems

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Suitable Grade Level

Grades 8-10