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The sum of an integer and 3 times the next consecutive even integer is -10 . Find the value of the greater integer. Answer:

The sum of an integer and 33 times the next consecutive even integer is −10-10 . Find the value of the greater integer.\newlineAnswer:

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Q. The sum of an integer and 33 times the next consecutive even integer is −10-10 . Find the value of the greater integer.\newlineAnswer:
  1. Define First Integer: Let's denote the first integer as xx. Since we are dealing with consecutive even integers, the next consecutive even integer would be x+2x + 2 (because even integers are 22 units apart).\newlineThe problem states that the sum of the first integer and three times the next consecutive even integer is −10-10. We can write this as an equation:\newlinex+3(x+2)=−10x + 3(x + 2) = -10
  2. Write Equation: Now, let's distribute the 33 into the parentheses: x+3x+6=−10x + 3x + 6 = -10
  3. Distribute 33: Combine like terms: 4x+6=−104x + 6 = -10
  4. Combine Like Terms: Subtract 66 from both sides to isolate the term with xx: \newline4x=−10−64x = -10 - 6\newline4x=−164x = -16
  5. Isolate xx Term: Divide both sides by 44 to solve for xx:x=−164x = \frac{-16}{4}x=−4x = -4
  6. Solve for x: Now that we have the value of the first integer x=−4x = -4, we need to find the next consecutive even integer, which is x+2x + 2:−4+2=−2-4 + 2 = -2So, the greater integer is −2-2.

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