Math, asked by krishna3651, 10 hours ago

Ahmad thinks of a number. He adds 1 then multiplies the result by 3. The answer is the same as 4 times the number take away 4. What number did Ahmad think of? (05) Solutio​

Answers

Answered by llMissCrispelloll
4

Answer:

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Let x be the number that Vincent thinks of originally. Adding 1 then multiplying the result by 3 would yield 3(x+1). But, this is the same as 4 times the number take away 4 (i.e. 4x-4).

 

Now, 3(x+1) = 4x - 4

 

3x + 3 = 4x - 4

 

Add 4 to both sides while also subtracting 3x will yield 7 = x.

 

Vincent's original number was 7.  To check add 1 (answer =8), then multiply it by 3 (answer = 24). Compare this with 4 times 7 (answer = 28) then take away 4 (answer = 24). Since they are both 24, then 7 is correct.

Answered by llMissCrispelloll
1

Answer:

\huge\underline{\overline{\mid{\bold{\mathbb{\purple{AnsWer}\mid}}}}}

Let x be the number that Vincent thinks of originally. Adding 1 then multiplying the result by 3 would yield 3(x+1). But, this is the same as 4 times the number take away 4 (i.e. 4x-4).

 

Now, 3(x+1) = 4x - 4

 

3x + 3 = 4x - 4

 

Add 4 to both sides while also subtracting 3x will yield 7 = x.

 

Vincent's original number was 7.  To check add 1 (answer =8), then multiply it by 3 (answer = 24). Compare this with 4 times 7 (answer = 28) then take away 4 (answer = 24). Since they are both 24, then 7 is correct.

Answered by llMissCrispelloll
1

Answer:

\huge\underline{\overline{\mid{\bold{\mathbb{\purple{AnsWer}\mid}}}}}

Let x be the number that Vincent thinks of originally. Adding 1 then multiplying the result by 3 would yield 3(x+1). But, this is the same as 4 times the number take away 4 (i.e. 4x-4).

 

Now, 3(x+1) = 4x - 4

 

3x + 3 = 4x - 4

 

Add 4 to both sides while also subtracting 3x will yield 7 = x.

 

Vincent's original number was 7.  To check add 1 (answer =8), then multiply it by 3 (answer = 24). Compare this with 4 times 7 (answer = 28) then take away 4 (answer = 24). Since they are both 24, then 7 is correct.

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