The sum of three consecutive odd integers. Given: The sum of the first two integers is equal to twenty-four less than four times the third integer, x+(x+2)=4(x+4)−24

Question

The sum of three consecutive odd integers.  Given: The sum of the first two integers is equal to twenty-four less than four times the third integer, x+(x+2)=4(x+4)−24

intacte87 · Accepted Answer

Calculation:  The sum of the first two integers is equal to twenty-four less than four times the third integer. x+(x+2)=4(x+4)−24. first simplify the like terms and then subtract both sides by 2x and then add both sides by 8 and simplify further divide both sides by 2 as shown below,  x+(x+2)=4(x+4)−24  x+(x+2)=4x+16−24  2x+2=4x−8  2x+2−2x=4x−8−2x  2=2x−8  2+8=2x−8+8  10=2x  102=2x2  5=x  ⇒x=5  Thus, the integers and their sum is 5,7,9 and sum 21.

Travis Hicks · Accepted Answer

Step 1  x+(x+2)=4(x+4)−24  Combine x and x to get 2x.  2x+2=4(x+4)−24  Use the distributive property to multiply 4 by x+4.  2x+2=4x+16−24  Subtract 24 from 16 to get −8.  2x+2=4x−8  Subtract 4x from both sides.  2x+2−4x=−8  Combine 2x and −4x to get −2x.  −2x+2=−8  Subtract 2 from both sides.  −2x=−8−2  Subtract 2 from −8 to get −10.  −2x=−10  Divide both sides by −2.  x=−10−2  Divide −10 by −2 to get 5.  x=5

The sum of three consecutive odd integers. Given: The sum of

Answered question

Answer & Explanation

New Questions in Pre-Algebra