1. ## Algebraic

1. The sum of 5 and x divided by the difference of x and 8 and the result is 5.

2. The sum of a number and 7 is equal to twice the difference of the number and 1.

3. Six more than six times a number is 24.

I can't seem to make any of these problems work out corectly.

2. Originally Posted by Patience
1. The sum of 5 and x divided by the difference of x and 8 and the result is 5.

2. The sum of a number and 7 is equal to twice the difference of the number and 1.

3. Six more than six times a number is 24.

I can't seem to make any of these problems work out corectly.
1. The sum of 5 and x divided by the difference of x and 8 and the result is 5.

Remember "sum" means add and "difference" means subtract. So we have
(x + 5)/(x - 8) = 5
=> x + 5 = 5(x - 8) ...........................I multiplied both sides by (x - 8)
=> x + 5 = 5x - 40 ............................I expanded the brackets
=> 40 + 5 = 5x - x ............................I put all the x's on one side and all the constants on the other
=> 45 = 4x
=> x = 45/4

2. The sum of a number and 7 is equal to twice the difference of the number and 1.

Note: sum = add, difference means subtract, twice means 2 times

Let our number be x. We have:

x + 7 = 2(x - 1)
=> x + 7 = 2x - 2
=> 7 + 2 = 2x - x
=> x = 9

3. Six more than six times a number is 24.

Let the number be x. We have

6x + 6 = 24
=> 6x = 24 - 6
=> 6x = 18
=> x = 18/6
=> x = 3

If there's anything you don't understand, say so

3. Thank you,

The problem I have most with algebraic is translating it to an equation.
You make it look so easy. The step by step you have showed me has help me so far.

4. Originally Posted by Patience
Thank you,

The problem I have most with algebraic is translating it to an equation.
You make it look so easy. The step by step you have showed me has help me so far.
So you understand how to appraoch these problems now? How about you post another question and it's solution to let me see if you get it.