2 consecutive integers such that 7 times the smaller is less than 6 times the greater. What are the greatest such intergers

Answer:

The numbers are

22 and 23  to solve a problem like this, we need to read and define as we go. Let me explain. So we know that there are two consecutive integers. They can be

x and x+1

. Since their consecutive, one has to be 1

number higher (or lower) than the other. Ok, so first we need "seven times the larger"( 7x+1)

Next, we need to "minus three times the smaller" 7(x+1)−3x

Is equal to " 957(x+1)−3x=95

Alright! There's the equation, now we just need to solve for  x! First we are going to get everything on one side and distribute the 7=7x+7−3x−95=4x−88

Pull out a

=4(x−22)Now that we have two terms, we can set them both equal to

0 and solve.4≠0

This can never be true, lets move to the next term (x−22)=0x=

22 That's it! So your two consecutive numbers are

22

The greatest such intergers are;  Lesser = x = 6. Greater = x + 1 = 7

To find the  sum of consecutive integers, we will use

[tex]1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}[/tex]

Here, Two consecutive integers represented as variables:

x, x + 1

Greater one: x + 1

Lesser one: x

Write as an equation

6(Greater) = 7(Lesser)

Substitute in known value;

6(x + 1) = 7(x)

Multiply

6x + 6 = 7x

Subtract 6x from both sides of the equation

6 = x

Lesser = x = 6

Greater = x + 1 = 6 + 1 = 7

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