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Zorluk: ZorSolving Linear Equations

A chemist is preparing a mixture. The volume of acid, VV in liters, required for a specific reaction satisfies the equation:

35(2V−7)−12(V+4)=110(3V−2)\frac{3}{5}(2V - 7) - \frac{1}{2}(V + 4) = \frac{1}{10}(3V - 2)

What is the value of the expression 4V+34V + 3?

  1. A
    15
  2. B
    23
  3. C
    27
  4. D
    47
  5. 63Cevap

Cevap

The value of the expression is 63.
The correct answer is 63. Multiplying both sides of the equation by the least common denominator of 10 eliminates the fractions and yields 6(2V−7)−5(V+4)=3V−26(2V - 7) - 5(V + 4) = 3V - 2. Distributing the factors results in 12V−42−5V−20=3V−212V - 42 - 5V - 20 = 3V - 2, which simplifies to 7V−62=3V−27V - 62 = 3V - 2. Subtracting 3V3V and adding 62 to both sides produces 4V=604V = 60, which gives V=15V = 15. Substituting 15 into the expression 4V+34V + 3 results in 4(15)+3=634(15) + 3 = 63.

Adım Adım Çözüm

1
Multiply the entire equation by the least common denominator (LCD) to eliminate the fractions.
6(2V−7)−5(V+4)=1(3V−2)6(2V - 7) - 5(V + 4) = 1(3V - 2)
The denominators are 5, 2, and 10, so the LCD is 10.
2
Distribute the constants on the left side of the equation.
12V−42−5V−20=3V−212V - 42 - 5V - 20 = 3V - 2
Distributing 6 to (2V−7)(2V - 7) yields 12V−4212V - 42, and distributing −5-5 to (V+4)(V + 4) yields −5V−20-5V - 20.
3
Combine like terms on the left side of the equation.
7V−62=3V−27V - 62 = 3V - 2
Combining the variable terms gives 12V−5V=7V12V - 5V = 7V, and combining the constant terms gives −42−20=−62-42 - 20 = -62.
4
Isolate the variable term by performing inverse operations.
4V=604V = 60
Subtract 3V3V from both sides to get 4V4V, and add 62 to both sides to get 60.
5
Solve for VV by dividing both sides of the equation.
V=15V = 15
Dividing 60 by 4 yields 15.
6
Substitute the value of VV into the requested expression.
4(15)+3=634(15) + 3 = 63
The question asks for the value of the expression 4V+34V + 3, not the value of VV.

Anahtar Kavram

Solving multi-step linear equations containing fractions by finding a common denominator, distributing terms correctly, and evaluating algebraic expressions.
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Solving Linear Equations

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