Question

the denominator of a rational number is greater than its numerator by 7 if 3 is subtracted from the numerator and 2 is added to the numerator the new number 1 by 5 find the rational number​

Answer 1

Answer:

Step-by-step explanation:

Given :-

The denominator of a rational number is greater than its numerator by 7.

If 3 is subtracted from the numerator and 2 is added to the numerator the new number 1 by 5.

To Find :-

The rational number

Solution :-

Let the numerator be x.

Denominator of the rational number = x + 7

The rational number = x/x+7 ..... (1)

According to the Question,

Therefore, x-3/(x+7)+2 = 1/5

⇒ x - 3/x + 9 = 1/5

⇒ 5x - 15 = x + 9

⇒ 5x - x = 9 + 1

⇒ 4x = 24

⇒ x = 24/4

⇒ x = 6

Putting x = 6 in (1), we get

Required rational number = 6/6+7

Required rational number = 6/13

Answer 2

$\bold{\huge{\underline{\underline{\mathfrak{AnsWer:}}}}}$

$\bold{\boxed{\large{\sf{\pink{Rational\:Number\:=\:{\dfrac{x}{y}}\:=\:{\dfrac{6}{13}}}}}}}$

$\bold{\huge{\underline{\underline{\mathfrak{StEp\:by\:stEp\:explanation:}}}}}$

GIVEN :

• The denominator of a rational number is greater than its numerator by 7
• If 3 is subtracted from the numerator and 2 is added to the denominator the new number $\bold{\dfrac{1}{5}}$

TO FIND :

• The rational number

SOLUTION :

Let the numerator of the rational number be x.

Let the denominator of the rational number be y.

Rational Number = $\bold{\dfrac{x}{y}}$

$\bold{\underline{\underline{\sf{As\:per\:the\:first\:given\:condition:}}}}$

• The denominator of a rational number is greater than its numerator by 7

Constituting it mathematically,

$\rightarrow$ $\bold{y=x+7}$

$\rightarrow$ $\bold{x+7=y}$

$\rightarrow$ $\bold{x-y=-7}$ --->(1)

$\bold{\underline{\underline{\sf{As\:per\:the\:second\:given\:condition:}}}}$

• If 3 is subtracted from the numerator and 2 is added to the denominator the new number $\bold{\dfrac{1}{5}}$

Numerator = x - 3

Denominator = y + 2

Rational number = $\bold{\dfrac{1}{5}}$

Constituting it mathematically,

$\rightarrow$ $\bold{\dfrac{x-3}{y+2}}$ = $\bold{\dfrac{1}{5}}$

Cross multiplying,

$\rightarrow$ $\bold{5(x-3)=y+2}$

$\rightarrow$ $\bold{5x-15=y+2}$

$\rightarrow$ $\bold{5x-y=2+15}$ ---> (2)

Multiply equation 1 by 5,

$\rightarrow$x - y = - 7

$\rightarrow$ 5x - 5y = - 35 ---> (3)

Solve equation 2 and equation 3 simultaneously by elimination method.

Subtract equation 3 from equation 2,

5x - 5y = - 35 ---> (3)

- (5x - y = 17) ------> (2)

--------------------------

- 4y = - 52

$\rightarrow$ $\bold{y={\dfrac{-52}{-4}}}$

$\rightarrow$ $\bold{y={\dfrac{52}{4}}}$

$\rightarrow$ $\bold{y=13}$

Substitute y = 13 in equation 1,

$\rightarrow$x - y = - 7

$\rightarrow$x - 13 = - 7

$\rightarrow$ x = - 7 + 13

$\rightarrow$ x = 6

$\bold{\boxed{\large{\sf{\red{Numerator\:=\:x\:=\:6}}}}}$

$\bold{\boxed{\large{\sf{\red{Denominator\:=\:y\:=\:13}}}}}$

$\bold{\boxed{\large{\sf{\red{Rational\:Number\:=\:{\dfrac{x}{y}}\:=\:{\dfrac{6}{13}}}}}}}$