35,000 views
8 votes
8 votes
Find two consecutive odd integers such that twice the greater is 17 more than the lesser.

User Dan Wagner
by
3.0k points

1 Answer

16 votes
16 votes

Answer:

13 and 15.

Explanation:

Let's set this as a system on linear equations.

Say the greater numbers is "x", and the lesser is "y".

First requirement:

The difference between the numbers must be 2, the problem states this. Therefore, we can state this:

x=y+2

This makes the difference between our two numbers equal 2.

Now, the second requirement is that "twice the greater is 17 more than the lesser". So, this is what should happen:

2x=17+y

Twice x, the greater, is 17 more than y, the lesser.

This is the resulting system on linear equations:

1. x=y+2

2. 2x=17+y

• Let's solve using the substitution method:

Step 1. Take equation 1 and solve for any of thhe varaibles, let's solve for x.

x= y+2

Step 2. Use this expression to substitute into equation 2.

2(y+2)=17+y

Step 3. Simplify.

2y+4=17+y

Step 4. Reorganize.

2y-y=17-4

Step 5. Simplify.

y= 13

Step 6. Take the value of y and apply it to equation 1 to find the value of x.

x=(13)+2

x= 15

• Now we have found our 2 numbers:

13

15

• Verification.

1. The difference between the numbers is 2: Correct.

2. The greater number's double is 17 more than the lesser:

15*2= 30

30-13= 17

Correct.

User Kevin Ross
by
2.3k points