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- WORD PROMBLEMS ON SIMULTANEOUS LINEAR EQUATIONS

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Sorry , but I could not understand the 2 and 5 answer ! It will be a great flavor if you solve it showing the algebra .

2.IF 2 IS ADDED TO THE NUMERATOR OF A FRACTION AND 1 IS SUBTRACTED FROM ITS DENOMINATOR , THE FRACTION BECOMES 4 BY 5 .IF 1 IS ADDED TO THE DENOMINATOR OF THE FRACTION , THE FRACTION BECOMES 1 BY 2 . FIND THE FRACTION.

5 . A. 2-DIGIT NO. OS 5 TIMES THE SUM OF ITS DIGITS. IF 9 IS ADDED TO THE NO. , ITS DIGITS ARE REVERSED . FIND THE NO.

2. The fraction is A/B. The problem says if we take (A+2)/(B-1), we get 4/5.

If we take A/(B+1), we get 1/2. This gives (A+2)/(B-1) = 4/5 and A/(B+1) = 1/2.

Using the 2nd equation and cross multiplying gives 2A = B+1, which is the same as B = 2A-1.

Putting this in the 1st equation gives (A+2)/(2A-1 - 1) on the left,

and this makes the equation (A+2)/(2A-2) = 4/5.

Cross multiplying gives 5A + 10 = 8A - 8.

Adding -5A + 8 to both sides gives 18 = 3A, so A = 6.

Looking back at equation 2 gives 6/(B+1) = 1/2.

Cross multiplying gives B+1 = 12, so B = 11.

Checking these values out gives (6+2)/(11-1) = 8/10 = 4/5, as given, and

6/(11+1) = 6/12 = 1/2, as given.

5. Let the two digit number be 10A + B.

This makes A the ten's digit and B the one's digit.

If 9 is added, the number is 10A + B + 9, and this reverses the digits.

This means the answer is 10B + A.

We then need to solve 10A + B + 9 = 10B + A.

Adding -A-B to both sides gives 9A + 9 = 9B.

Dividing by 9 gives A + 1 = B.

Now for this, almost any A works, for is A=1, then B=2, and 21 = 12 + 9;

If A=2, then B=3, and 32 = 23 + 9; is A=5, the B=5, and 54 = 45 + 9.

Looking back at the 1st condition, it says that 5(A+B) = 10A + B.

That is the same as 5A + 5B = 10A + B. Adding -5A - B to both sides gives 4B = 5A.

Since A and B are single digits, this means B = 5 and A = 4, so we get 20 = 20.

This makes the number 45, for that is five times the sum of the digits and

adding 9 reverses the digits.

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