Asked by anonymous
How do I express this in algebraic form? And could you please explain how you came up with the expression?
"Come up with a number. Add 3 to it. Muliply by 1 more than the original number. Subtract 3. Divide by the original number. Subtract 4. You should get the original number.
How it works: x is the original number. Add 3 to x to get (x+3). Multiply by (x+1) to get x² + 4x + 3. Subtract 3 to get x² + 4x, which factorises to x(x+4). Dividing by the original number, x, gives x + 4, and subtracting 4 gives the original number, x."
"Come up with a number. Add 3 to it. Muliply by 1 more than the original number. Subtract 3. Divide by the original number. Subtract 4. You should get the original number.
How it works: x is the original number. Add 3 to x to get (x+3). Multiply by (x+1) to get x² + 4x + 3. Subtract 3 to get x² + 4x, which factorises to x(x+4). Dividing by the original number, x, gives x + 4, and subtracting 4 gives the original number, x."
Answers
Answered by
anonymous
Is (x^2+4x/x)-4 correct?
Answered by
Reiny
Come up with a number: x
Add 3 to it : x+3
Muliply by 1 more than the original number : (x+3)(x+1)
Subtract 3 : (x+3)(x+1) - 3
Divide by the original number : [(x+3)(x+1) - 3]/x
Subtract 4 : [(x+3)(x+1) - 3]/x - 4
<b>[(x+3)(x+1) - 3]/x - 4</b>
= (x^2 + 4x +3 - 3)/x - 4
= (x^2 + 4x)/x - 4
= x + 4 - 4
= x
Notice the long expression follows the actual wording.
Your answer of (x^2+4x/x)-4 needs brackets like this:
(x^2+4x)/x - 4 or else only the 4x is divided by x
Add 3 to it : x+3
Muliply by 1 more than the original number : (x+3)(x+1)
Subtract 3 : (x+3)(x+1) - 3
Divide by the original number : [(x+3)(x+1) - 3]/x
Subtract 4 : [(x+3)(x+1) - 3]/x - 4
<b>[(x+3)(x+1) - 3]/x - 4</b>
= (x^2 + 4x +3 - 3)/x - 4
= (x^2 + 4x)/x - 4
= x + 4 - 4
= x
Notice the long expression follows the actual wording.
Your answer of (x^2+4x/x)-4 needs brackets like this:
(x^2+4x)/x - 4 or else only the 4x is divided by x
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