Algebra 1 math problem

Question

Find two consecutive integers such that ten more than twice the smaller is seven less than three times the larger

Answer 1

Let x be the smaller integer.  Since they are consecutive the larger integer is x + 1.

 

Now 2x + 10 = 3(x + 1) - 7

2x + 10 = 3x + 3 - 7 = 3x -4

3x - 2x = 10 - (-4)

x = 10 + 4 = 14

 

The integers are 14 and 15.

Answer 2

Let the consecutive integers be:

 

n

n + 1

 

n is the smaller, n + 1 is the larger.

 

The given information translates into:

 

2n + 10  [10 more than twice the smaller]

 

3(n + 1) - 7  [7 less than 3 times the lager]

 

Then

 

2n + 10 = 3(n + 1) - 7

 

Solve for n.

 

2n + 10 = 3n + 3 - 7

 

10 - 3 + 7 = 3n - 2n

 

14 = n

n+1 = 15

 

The two integers are 14 and 15.  Go ahead and test the solutions to see that they satisfy the given conditions.