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If 6 is subtracted from the third of three consecutive odd integers and the result is multiplied by 2, the answer is 23 less then the sun if the first and twice the second of the integers

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\begin{gathered} \text{Let }x\text{ be the first odd integer, then it goes that} \\ x+2\rightarrow\text{second odd integer} \\ x+4\rightarrow\text{third odd integer} \\ \text{If 6 is subtracted from the third of three consecutive odd integer, translates to} \\ (x+4)-6\Rightarrow x+4-6\Rightarrow x-2 \\ \text{and the result is multiplied by 2, which translates to} \\ (x-2)\cdot2\Rightarrow2x-4 \\ \text{the answer is 23 less than the sum of the first and twice of the second integers} \\ (x)+2(x+2)\rightarrow x+2x+4\rightarrow3x+4 \\ (3x+4)-23 \\ 3x-19 \\ \text{equate them both sides and we have} \\ 2x-4=3x-19,\text{ left hand is the first part, right hand is the second part} \\ \text{solve for }x \\ 2x-4=3x-19 \\ -4+19=3x-2x \\ 15=x \\ x=15 \\ \text{so the three odd integers are, } \\ x=15 \\ x+2=17 \\ x+4=19 \\ 15,17,19 \end{gathered}

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