Anonymous

# Applying equations to solving word problems (grade nine)?

problems like:

a) Find five consecutive integers such that the sum of the first and fourth is 533

b) Find four consecutive integers if the sum of the third and fourth is 63

Im kind of confused as to how to do these so can someone explain it to me???

Thanks

Relevance

An integer is simply a whole number between negative infinity and positive infinity. The first integer in the set can be named x, and just add 1 to each additional integer needed as long as they are in sequence.

I'll solve the first one for you.

You want the first integer (we will call this x, because we don't know what it is) and the fourth integer (we will call this x + 3, because we don't know what it is, but we do know that it is 3 more than x) to add up to 533. You can determine the integers in between these two after you have determined x.

So, our equation is: 533 = x + (x+3)

We can drop the paranthesis because no matter what order we add them in, we will come to the same number.

So now we have: 533 = x + x + 3

This simplifies to: 533 = 2x + 3

Now we subtract 3 from the right side and left side to start isolating x:

533 (- 3) = 2x + 3 (-3)

This gives us: 530 = 2x

Divide both sides by 2 to isolate x on the right side:

530/2 = 2x/2

This leaves us with: 265 = x

So we have determined that x (which, remember, is the first integer in the series) is 265. So the second number is 266, the third is 267, and the fourth is 268.

Add the first integer and the fourth integer together to check and make sure we got the right answer: is 265 + 268 = 533?

265 + 268 = 533.

The first and third integer add up to 533, so we know we have the correct value for x.

• Anonymous

a) Find five consecutive integers such that the sum of the first and fourth is 533

x + (x + 3) = 533

2x + 3 = 533

2x = 530

x = 265

So if the first integer is 265, then the five consecutive are:

1: 265

2: 266

3: 267

4: 268

5: 269

Adding the first and fourth: 265 + 268 = 533

b) Find four consecutive integers if the sum of the third and fourth is 63

Let the first integer = x, then the third is x + 2 and the fourth is x + 3

(x + 2) + (x + 3) = 63

2x + 5 = 63

2x = 58

x = 29

So the four consecutive integers are:

1: 29

2: 30

3: 31

4: 32

And the sum of the third and fourth is: 31 + 32 = 63

Hope that helps. :)

• a) let x = first

let (x+1) = 2nd

let (x+2) = 3rd

let (x+3) = 4th

let (x+4) = 5th

Add 1st and 4th = 533

x + x+3 = 533

2x + 3 = 533

2x = 530

x = 265

So the integers are 265, 266, 267, 268, 269

Use that to try b)

• Anonymous

a)x+(x+4)= 533

2x+4=533

Subtract 4 on both sides.

2x=529

Divide by 2 on both sides.

x=264.5

The numbers are 264.5 and 268.5.

b)(x+3)+(x+4)=63

2x+7=63

Subtract 7 on both sides.

2x=56

Divide by 2 on both sides.

x=28

The numbers are 31 and 32.