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## Math Algebra 2 Help

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### Solved Examples

**Question 1:**Seven times a two digits number is equal to four times the number obtained by reversing the order of digits and sum of the digits of the number is 3. Find the number ?

**Solution:**

Step 1:

Let the digit at ten's place = x and digit at unit's place = y

=> Number = 10x + y

Number obtained on reversing the digits = 10y + x

Sum of digits = x + y

Step 2:

Case 1:

Seven times a two digits number is equal to four times the number obtained by reversing the order of digits.

7(10x + y) = 4(10y + x) .........................................(1)

Case 2:

Sum of the digits of the number is 3.

x + y = 3 ...........................................(2)

Step 3:

(1) => 70x + 7y = 40y + 4x

=> 70x - 4x = 40y - 7y

=> 66x = 33y

Divide each side by 33

=> 2x = y .......................................(3)

Step 4:

Put y = 2x in equation (2)

=> x + 2x = 3

=> 3x = 3

=> x = 1, put in (3)

=> 2 * 1 = y

=> y = 2

=> x = 1 and y = 2

Let the digit at ten's place = x and digit at unit's place = y

=> Number = 10x + y

Number obtained on reversing the digits = 10y + x

Sum of digits = x + y

Step 2:

Case 1:

Seven times a two digits number is equal to four times the number obtained by reversing the order of digits.

7(10x + y) = 4(10y + x) .........................................(1)

Case 2:

Sum of the digits of the number is 3.

x + y = 3 ...........................................(2)

Step 3:

(1) => 70x + 7y = 40y + 4x

=> 70x - 4x = 40y - 7y

=> 66x = 33y

Divide each side by 33

=> 2x = y .......................................(3)

Step 4:

Put y = 2x in equation (2)

=> x + 2x = 3

=> 3x = 3

=> x = 1, put in (3)

=> 2 * 1 = y

=> y = 2

=> x = 1 and y = 2

**Answer**: Original number is 12.**Question 2:**A fraction becomes $\frac{3}{5}$ if 1 added to both numerator and denominator. If 5 subtracted from both numerator and denominator, the fraction becomes $\frac{2}{4}$. What is the fraction.

**Solution:**

Step 1:

Let the fraction be $\frac{x}{y}$

Case 1:

$\frac{x + 1}{y + 1} = \frac{3}{5}$ ..............................(1)

Case 2:

$\frac{x - 5}{y - 5} = \frac{2}{4}$ = $\frac{1}{2}$ .................................(2)

Step 2:

(1) => 5(x + 1) = 3(y + 1)

=> 5x + 5 = 3y + 3

=> 5x - 3y + 2 = 0 ......................................(3)

and (2) =>

2(x - 5) = y - 5

=> 2x - 10 = y - 5

=> 2x - 5 = y .........................................(4)

Step 3:

Put y = 2x - 5 in equation (3)

=> 5x - 3( 2x - 5) + 2 = 0

=> 5x - 6x + 15 + 2 = 0

=> - x + 17 = 0

=> x = 17

Step 4:

Put x = 17 in equation (4)

=> 2 * 17 - 5 = y

=> 34 - 5 = y

=> 29 = y

Let the fraction be $\frac{x}{y}$

Case 1:

$\frac{x + 1}{y + 1} = \frac{3}{5}$ ..............................(1)

Case 2:

$\frac{x - 5}{y - 5} = \frac{2}{4}$ = $\frac{1}{2}$ .................................(2)

Step 2:

(1) => 5(x + 1) = 3(y + 1)

=> 5x + 5 = 3y + 3

=> 5x - 3y + 2 = 0 ......................................(3)

and (2) =>

2(x - 5) = y - 5

=> 2x - 10 = y - 5

=> 2x - 5 = y .........................................(4)

Step 3:

Put y = 2x - 5 in equation (3)

=> 5x - 3( 2x - 5) + 2 = 0

=> 5x - 6x + 15 + 2 = 0

=> - x + 17 = 0

=> x = 17

Step 4:

Put x = 17 in equation (4)

=> 2 * 17 - 5 = y

=> 34 - 5 = y

=> 29 = y

**Answer:**The fraction is $\frac{17}{29}$.**Question 3:**Find the LCM of 18, 42 and 34.

**Solution:**

Prime Factors of 18 = 2 x 3 x 3

Prime Factors of 42 = 2 x 3 x 7

Prime Factors of 34 = 2 x 17

=> LCM(18, 34, 42) = 2 x 3 x 3 x 7 x 17

= 2142

=> LCM(18, 34, 42) = 2142

Prime Factors of 42 = 2 x 3 x 7

Prime Factors of 34 = 2 x 17

=> LCM(18, 34, 42) = 2 x 3 x 3 x 7 x 17

= 2142

=> LCM(18, 34, 42) = 2142