Math Problem Statement

1/2(3/2y+1/3)=-1/2(y-5/2)

Solution

Let's solve the equation step by step:

Given equation: 12(32y+13)=12(y52)\frac{1}{2} \left(\frac{3}{2} y + \frac{1}{3} \right) = -\frac{1}{2} \left( y - \frac{5}{2} \right)

Step 1: Eliminate the fractions by multiplying both sides by 2.

(32y+13)=(y52)\left(\frac{3}{2} y + \frac{1}{3} \right) = - ( y - \frac{5}{2} )

Step 2: Distribute on both sides.

On the left-hand side, nothing changes: 32y+13\frac{3}{2} y + \frac{1}{3} On the right-hand side, distribute the negative sign: y+52- y + \frac{5}{2} So now we have: 32y+13=y+52\frac{3}{2} y + \frac{1}{3} = - y + \frac{5}{2}

Step 3: Eliminate the fractions by multiplying the entire equation by 6 (the least common denominator of 2 and 3).

Multiplying everything by 6: 6×(32y)+6×(13)=6×(y)+6×(52)6 \times \left( \frac{3}{2} y \right) + 6 \times \left( \frac{1}{3} \right) = 6 \times \left( - y \right) + 6 \times \left( \frac{5}{2} \right) Simplify: 9y+2=6y+159y + 2 = -6y + 15

Step 4: Solve for yy.

Move all terms involving yy to one side and constant terms to the other: 9y+6y=1529y + 6y = 15 - 2 15y=1315y = 13 Now, solve for yy: y=1315y = \frac{13}{15}

Thus, the solution is: y=1315y = \frac{13}{15}

Would you like more details, or do you have any questions?

Related Questions:

  1. How can we simplify equations with fractions faster?
  2. What is the method to solve equations that have variables on both sides?
  3. What are the steps to solve linear equations with fractions?
  4. How do you find the least common denominator (LCD) in an equation?
  5. What are some common mistakes to avoid when solving linear equations?

Tip:

When solving equations with fractions, multiplying through by the least common denominator (LCD) can help eliminate the fractions early, making the equation easier to handle.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Fractions
Distributive Property

Formulas

Distributive Property: a(b + c) = ab + ac

Theorems

Distributive Property
Simplifying Fractions

Suitable Grade Level

Grades 8-10