Answers:
- A+B+C = 7x - 9y - z
- A-B+C = 3x - 3y - 9z
- A+B-C = -x + y + 11z
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Explanation:
Skip to the "Work Shown" section below to see the steps for each part.
Each variable is a place holder for some unknown number. We can think of it like a box that holds a number. Luckily the letter x is found in the word "box" to help cement this line of thinking.
1x = 1 box
2x = 2 boxes
3x = 3 boxes
and so on
This shorthand notation allows us to quickly add boxes
2 boxes + 3 boxes = 5 boxes
So 2x+3x = 5x
And we can subtract boxes as well
7 boxes - 3 boxes = 7x - 3x = 4x = 4 boxes
The same number must go in box x for this to work out.
When it comes to dealing with negative numbers, it might help to use a vertical number line. I like to think of it like going up and down an elevator on a skyscraper. Negative integers represent basement levels.
It's possible to have boxes within boxes. The box A has the three boxes x-y+z inside it. So wherever we see an "A", replace it with x-y+z. Unfortunately we don't know what numbers go in the boxes for x, -y and z. Use the same type of thinking for B and C.
Don't forget to use the distributive property to distribute the negative through to each term.
Example: (x+y)-(w+z) = x+y-w-z
Many students make the mistake in saying that (x+y)-(w+z) = x+y-w+z which is not true. The +z should be -z at the very end.
Let me know if you have any other questions.
The next section below shows the steps needed to get the three answers.
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Work Shown:
Part (I)
A+B+C = (x-y+z)+(2x-3y+4z)+(4x-5y-6z)
A+B+C = (x+2x+4x)+(-y-3y-5y)+(z+4z-6z)
A+B+C = 7x - 9y - z
Part (II)
A-B+C = (x-y+z)-(2x-3y+4z)+(4x-5y-6z)
A-B+C = x-y+z-2x+3y-4z+4x-5y-6z
A-B+C = (x-2x+4x)+(-y+3y-5y)+(z-4z-6z)
A-B+C = 3x - 3y - 9z
Part (III)
A+B-C = (x-y+z)+(2x-3y+4z)-(4x-5y-6z)
A+B-C = x-y+z+2x-3y+4z-4x+5y+6z
A+B-C = (x+2x-4x)+(-y-3y+5y)+(z+4z+6z)
A+B-C = -x + y + 11z
