Algebraic Challenge: Simplify 2-3((-14):(5x/y)-y:(3x×2))

Order of Operations with Fraction Division

23((14):5xyy:(3x2))=? 2-3((-14):\frac{5x}{y}-y:(3x\cdot2))=\text{?}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:06 Division is also multiplication by the reciprocal
00:14 Let's write division as a fraction
00:23 Negative times negative always equals positive
00:28 Move the multiplication to the numerator
00:47 Simplify what we can
00:59 Find a common denominator, multiply each numerator by the other denominator
01:15 Combine the fractions into one fraction
01:23 Divide 89 by 10
01:26 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

23((14):5xyy:(3x2))=? 2-3((-14):\frac{5x}{y}-y:(3x\cdot2))=\text{?}

2

Step-by-step solution

To solve this problem, let's follow these steps:

  • Step 1: Simplify the innermost expression within parentheses
  • Step 2: Compute the division and subtraction
  • Step 3: Multiply the result by 3-3
  • Step 4: Add the result to 2

Now, let's work through each step:

Step 1: First, focus on (14):5xy(-14):\frac{5x}{y}. The operation suggests dividing 14-14 by 5xy\frac{5x}{y}, which is equivalent to multiplying by the reciprocal: 14×y5x-14 \times \frac{y}{5x}.

Step 2: Simplify the reciprocal multiplication, 14×y5x=14y5x-14 \times \frac{y}{5x} = -\frac{14y}{5x}.

Step 3: Now consider the other term y:(3x2)=y6xy:(3x \cdot 2) = \frac{y}{6x}.

Step 4: Subtract these results: 14y5xy6x-\frac{14y}{5x} - \frac{y}{6x}. To combine fractions, find a common denominator (30x):

14y5x=84y30x-\frac{14y}{5x} = -\frac{84y}{30x} and y6x=5y30x\frac{y}{6x} = \frac{5y}{30x}.

Step 5: The subtraction becomes 84y30x5y30x=89y30x-\frac{84y}{30x} - \frac{5y}{30x} = -\frac{89y}{30x}.

Step 6: Multiply this result by 3-3: 3×(89y30x)=267y30x3 \times \left(-\frac{89y}{30x}\right) = \frac{267y}{30x}. Simplify fractionally to get 8.9yx\frac{8.9y}{x}.

Step 7: Add 2 to this result: 2+8.9yx2 + \frac{8.9y}{x}.

Therefore, the solution to the problem is 2+8.9yx 2 + 8.9\frac{y}{x} .

3

Final Answer

2+8.9yx 2+8.9\frac{y}{x}

Key Points to Remember

Essential concepts to master this topic
  • Rule: Division by a fraction equals multiplication by its reciprocal
  • Technique: Convert (14):5xy (-14):\frac{5x}{y} to 14×y5x -14 \times \frac{y}{5x}
  • Check: Verify common denominators: 84y30x+5y30x=89y30x \frac{84y}{30x} + \frac{5y}{30x} = \frac{89y}{30x}

Common Mistakes

Avoid these frequent errors
  • Treating the colon (:) as simple division instead of fraction division
    Don't treat (-14):(5x/y) as -14/(5x/y) = wrong approach! This ignores that dividing by a fraction requires multiplying by its reciprocal. Always convert division by fractions to multiplication by the reciprocal first.

Practice Quiz

Test your knowledge with interactive questions

\( 70:(14\times5)= \)

FAQ

Everything you need to know about this question

What does the colon (:) symbol mean in this expression?

+

The colon means division, so (14):5xy (-14):\frac{5x}{y} means (14)÷5xy (-14) \div \frac{5x}{y} . When dividing by a fraction, flip it and multiply instead!

Why do I need to find a common denominator for the fractions?

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You can only add or subtract fractions with the same denominator. Since we have 14y5x \frac{14y}{5x} and y6x \frac{y}{6x} , we need common denominator 30x to combine them.

How did 267/30 become 8.9?

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Divide 267 by 30: 267 ÷ 30 = 8.9. This converts the improper fraction to a mixed decimal form that matches the answer choices.

Why is the final answer positive when we started with negative terms?

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We multiply the negative result 89y30x -\frac{89y}{30x} by -3, and negative times negative equals positive! So 3×(89y30x)=+267y30x -3 \times (-\frac{89y}{30x}) = +\frac{267y}{30x} .

What's the correct order of operations here?

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Follow PEMDAS:

  • First: Simplify inside parentheses (division and subtraction)
  • Second: Multiply by -3
  • Third: Add to 2

Always work from the innermost parentheses outward!

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