Math Problem Statement

Solve using the addition and multiplication principles together. Do not forget to check. one third y minus three fourths y plus 7 equals five twelfths plus one fourth y 1 3y− 3 4y+7= 5 12+ 1 4y

Solution

We are given the equation:

13y34y+7=512+14y\frac{1}{3}y - \frac{3}{4}y + 7 = \frac{5}{12} + \frac{1}{4}y

Step 1: Eliminate the fractions

To make things simpler, we will eliminate the fractions by multiplying the entire equation by the least common denominator (LCD) of the fractions. The denominators are 3, 4, and 12, and the LCD of these numbers is 12.

Multiply the entire equation by 12:

12×(13y34y+7)=12×(512+14y)12 \times \left( \frac{1}{3}y - \frac{3}{4}y + 7 \right) = 12 \times \left( \frac{5}{12} + \frac{1}{4}y \right)

This gives:

4y9y+84=5+3y4y - 9y + 84 = 5 + 3y

Step 2: Simplify both sides

Simplify the left and right sides:

(4y9y)+84=5+3y(4y - 9y) + 84 = 5 + 3y 5y+84=5+3y-5y + 84 = 5 + 3y

Step 3: Get all the yy-terms on one side

To move the yy-terms to one side, subtract 3y3y from both sides:

5y3y+84=5-5y - 3y + 84 = 5 8y+84=5-8y + 84 = 5

Step 4: Isolate the yy-term

Subtract 84 from both sides:

8y=584-8y = 5 - 84 8y=79-8y = -79

Step 5: Solve for yy

Divide both sides by -8:

y=798y = \frac{-79}{-8} y=798y = \frac{79}{8}

Step 6: Check the solution

Substitute y=798y = \frac{79}{8} back into the original equation:

13(798)34(798)+7=512+14(798)\frac{1}{3}\left(\frac{79}{8}\right) - \frac{3}{4}\left(\frac{79}{8}\right) + 7 = \frac{5}{12} + \frac{1}{4}\left(\frac{79}{8}\right)

The left-hand side simplifies to:

792423732+7\frac{79}{24} - \frac{237}{32} + 7

The right-hand side simplifies to:

512+7932\frac{5}{12} + \frac{79}{32}

Both sides are equal, so the solution y=798y = \frac{79}{8} is correct.

Would you like to go over any details or ask questions?

Here are 5 related questions for further exploration:

  1. How do you find the least common denominator (LCD) for fractions?
  2. How would you solve the equation if there were decimals instead of fractions?
  3. What would the solution be if the constants on the right-hand side were negative?
  4. How do we deal with equations that have no solutions?
  5. Can you solve this type of equation using substitution?

Tip: Always check your solution by plugging it back into the original equation to verify!

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Fractions
Simplification

Formulas

Multiplication Principle for Fractions: Multiply both sides by the least common denominator (LCD) to eliminate fractions.

Theorems

Addition and Multiplication Principles for Solving Equations

Suitable Grade Level

Grades 8-10