Question

If ( x + 4), (x+12), (x - 1),(x + 15) a are in
Q22
proportion then find the value of x​

Answer 1

Given :

(x + 4), (x+12), (x - 1),(x + 15) are in proportion.

To Find :

The value of x.

Solution :

Analysis :

Here it is said that the variables are in proportion. So we have to solve it by using the proportion laws.

Explanation :

a : b :: c : d

where,

• a = x + 4
• b = x + 12
• c = x - 1
• d = x + 15

$\\ :\implies\sf\dfrac{a}{b}=\dfrac{c}{d}$

Substituting the values of a, b, c, d;

$\\ :\implies\sf\dfrac{x+4}{x+12}=\dfrac{x-1}{x+15}$

By cross multiplying,

$\\ :\implies\sf(x+4)(x+15)=(x+12)(x-1)$

Expanding the brackets,

$\\ :\implies\sf x^2+15x+4x+60=x^2-x+12x-12$

Cancelling x² from both the sides,

$\\ :\implies\sf \cancel{x^2}\ +15x+4x+60=\cancel{x^2}\ -x+12x-12$

After evaluation,

$\\ :\implies\sf15x+4x+60=-x+12x-12$

$\\ :\implies\sf19x+60=11x-12$

Transposing 11x to LHS and 60 to RHS,

$\\ :\implies\sf19x-11x=-12-60$

After evaluation,

$\\ :\implies\sf8x=-72$

$\\ :\implies\sf x=\dfrac{-72}{\ \ 8}$

$\\ :\implies\sf x=\cancel{\dfrac{-72}{\ \ 8}}$

$\\ :\implies\sf x=-9$

$\\ \therefore\boxed{\bf x=-9.}$

The value of x is -9.

Verification :

LHS :

$\\ :\implies\sf\dfrac{x+4}{x+12}$

• Putting x = -9,

$\\ :\implies\sf\dfrac{(-9)+4}{(-9)+12}$

$\\ :\implies\sf\dfrac{-9+4}{-9+12}$

$\\ \therefore\bf\dfrac{-5}{3}$

RHS :

$\\ :\implies\sf\dfrac{x-1}{x+15}$

• Putting x = -9,

$\\ :\implies\sf\dfrac{(-9)-1}{(-9)+15}$

$\\ :\implies\sf\dfrac{-9-1}{-9+15}$

$\\ :\implies\sf\dfrac{-10}{6}$

$\\ :\implies\sf\dfrac{\cancel{-10}\ \ ^{-5}}{\not{6}\ \ ^3}$

$\\ \therefore\bf\dfrac{-5}{3}$

LHS = RHS.

• Hence verified.