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Difficulty: HardSolving One-Variable Linear Equations

Calculate the value of the unknown variable.

Answer:Calculate the value of kk that makes the following equation true:

34(8k−12)−25(15−5k)=3(k+2)+4 \frac{3}{4}(8k - 12) - \frac{2}{5}(15 - 5k) = 3(k + 2) + 4

k=k = 【5】

Answer

5
The correct value is 5. By distributing the fractions on the left, the expression becomes 6k−9−6+2k6k - 9 - 6 + 2k. Combining like terms simplifies this to 8k−158k - 15. On the right side, distributing the 3 results in 3k+6+43k + 6 + 4, which simplifies to 3k+103k + 10. Equating both simplified sides gives 8k−15=3k+108k - 15 = 3k + 10. Subtracting 3k3k from both sides leaves 5k−15=105k - 15 = 10, and adding 1515 to both sides yields 5k=255k = 25. Dividing by 5 results in k=5k = 5.

Step-by-Step Solution

1
Distribute the fractional terms on the left side of the equation.
6k−9−6+2k6k - 9 - 6 + 2k
Multiplying 34\frac{3}{4} by (8k−12)(8k - 12) yields 6k−96k - 9. Multiplying −25-\frac{2}{5} by (15−5k)(15 - 5k) yields −6+2k-6 + 2k. Paying close attention to the negative sign during distribution is critical.
2
Distribute the terms on the right side of the equation.
3k+6+43k + 6 + 4
Multiplying 33 by (k+2)(k + 2) expands the expression, setting up the right side for simplification.
3
Combine like terms on both sides of the equal sign.
8k−15=3k+108k - 15 = 3k + 10
On the left side, combine the variable terms (6k+2k=8k6k + 2k = 8k) and constant terms (−9−6=−15-9 - 6 = -15). On the right side, combine the constant terms (6+4=106 + 4 = 10).
4
Isolate the variable kk to find its value.
5k=255k = 25, which simplifies to k=5k = 5.
Subtract 3k3k from both sides to group the variables on the left (5k−15=105k - 15 = 10). Then, add 1515 to both sides to group the constants on the right (5k=255k = 25). Dividing both sides by 55 yields the final solution.

Key Concept

Solving One-Variable Linear Equations
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Topics

Solving One-Variable Linear Equations

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