find three consecutive integers such that 5 times the third equals twice the first plus seven more than the second

Question

find three consecutive integers such that 5 times the third equals twice the first plus seven more than the second

Answer 1

x,x+1, x+2

5(x+2)=2(x)+7+(x+1)

5X+10=3X+8

2X=-2

X=-1

-1,0,1

Answer 2

3 consecutive integers can be written as

x, x+1, and x+2

 

5 times the 3rd equals 2 times the 1st plus 7 more than the 2nd:

 

5(x+2) = 2x + x + 1 + 7  multiply thru the parenthesis and combine terms

5x + 5(2) = 3x + 8

5x + 10 = 3x + 8            subtract 3x from both sides

2x + 10 = 8                    subtract 10 from both sides

2x = -2                          divide both sides by 2

x = -1

 

Your consecutive integers are:

-1, -1+1, -1+2  =  -1, 0, 1

 

Check answer:  5(1) = 2(-1) + 0 + 7

                       5 = -2 + 7

                       5 = 5

Answer checks as correct

 

 

Answer 3

Three consecutive integers: x, x+1, and x+2

 

5 times the third => 5(x+2)  

twice the first => 2x

seven more than the second => (x+1)+7

 

Now, put the pieces together to match your statement:

 

5(x+2) = 2x + (x + 1) + 7

 

5x + 10 = 3x + 8                  Multiply (x + 2) by 5 and combine all like terms

 

5x + 10 = 3x + 8                  Subtract 3x from both sides

 

2x + 10 = 8                          Subtract 10 from both sides

 

2x = -2                                Divide both sides by 2

 

x = -1

 

 

Now go back to your consecutive integers: x, x+1, x+2

 

Substitute -1 in for x and solve:

 

x => -1

x+1 => -1 + 1 = 0

x+2 => -1 + 2 = 1

 

So your consecutive integers are -1,0, and 1.

 

Check your answer:

 

5(x+2) = 2x + (x + 1) + 7

5(1) = 2(-1) + 0 + 7

5 = -2 + 7

5 = 5