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Zorluk: ZorSolving One-Variable Linear Equations

Solve the following equation for pp:

23(6p−15)+3p=−3(p−8)+2p−10 \frac{2}{3}(6p - 15) + 3p = -3(p - 8) + 2p - 10

What is the final value of pp?

  1. A
    4
  2. B
    2
  3. 3Cevap
  4. D
    -3

Cevap

The correct value for the variable is 3.
The correct answer is 3. Distributing the terms on both sides yields 4p−10+3p=−3p+24+2p−104p - 10 + 3p = -3p + 24 + 2p - 10. Combining like terms results in 7p−10=−p+147p - 10 = -p + 14. Adding pp to both sides and adding 10 to both sides gives 8p=248p = 24. Dividing by 8 isolates the variable, resulting in p=3p = 3.

Adım Adım Çözüm

1
Distribute the coefficients to remove parentheses on both sides of the equation.
4p−10+3p=−3p+24+2p−104p - 10 + 3p = -3p + 24 + 2p - 10
To simplify the equation, all terms inside parentheses must be multiplied by their outer coefficients. Note that 23×6p=4p\frac{2}{3} \times 6p = 4p and 23×−15=−10\frac{2}{3} \times -15 = -10.
2
Combine like terms on each side of the equation independently.
7p−10=−p+147p - 10 = -p + 14
Combining variable terms and constant terms on the same side simplifies the equation structure before isolating the variable.
3
Move all variable terms to one side of the equation.
8p−10=148p - 10 = 14
Adding pp to both sides cancels the −p-p on the right side, grouping all unknowns on the left.
4
Isolate the variable by moving constants to the other side and dividing.
8p=24⇒p=38p = 24 \Rightarrow p = 3
Adding 10 to both sides moves the constant, and dividing both sides by the coefficient 8 solves for pp.

Anahtar Kavram

Solving complex one-variable linear equations requiring distribution and combining like terms.
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Konular

Solving One-Variable Linear Equations

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