Final answer:
After comparing each option with the linear equation 3x – y = 10 by algebraic manipulation, none of the provided options (a, b, c, or d) match the original equation. Therefore, none of the options are solutions to the given linear equation.
Step-by-step explanation:
To solve which options are solutions to the linear equation 3x – y = 10, we need to transform the given options into this form and see if they match or can be derived from the original equation. We will go through each option and perform necessary algebraic manipulations to determine if they can be solutions:
For a. y = 3x + 5, if we try to match it to the original equation, we would end up with 3x - (3x + 5) = 10, which simplifies to -5 = 10. This is not true, so option a is not a solution.
For b. y = –13x + 17, matching it to the original equation gives us 3x - (–13x + 17) = 10, which simplifies to 16x = -7. This does not correspond to our original equation, so option b is not a solution.
For c. x + 3y = 27, we can rewrite this as y = (27 - x) / 3. Multiplying both sides by 3 gives us 3y = 27 - x, or x + 3y = 27, which does not fit the form of the original equation, so option c is not a solution.
Finally, for d. y – 2 = 13(3x – 36), we expand the right side to get y - 2 = 39x - 468. Adding 2 to both sides gives y = 39x - 466, which does not match our original equation, so option d is also not a solution.
None of the provided options are solutions to the equation 3x – y = 10.