Question

The Algebraic equation is given as $\dfrac{{16}}{x} - 1 = \dfrac{{15}}{{x + 1}};x \ne 0, - 1,$ then one of the value of x is

Answer 1

Hint-In this question, we use the concept of cross multiplication to convert fraction into normal form and then use the basic property of algebra like multiply of two brackets $\left( {a + b} \right) \times \left( {c + d} \right) = a\left( {c + d} \right) + b\left( {c + d} \right)$.

Complete step-by-step solution -
Now, equation is \[\dfrac{{16}}{x} - 1 = \dfrac{{15}}{{x + 1}}\]
\[ \Rightarrow \dfrac{{16}}{x} - \dfrac{1}{1} = \dfrac{{15}}{{x + 1}}\]
Take a LCM between x and 1
\[ \Rightarrow \dfrac{{16 - x}}{x} = \dfrac{{15}}{{x + 1}}\]
Now, we convert fraction into normal form by using cross multiplication.
Apply cross multiplication,
\[ \Rightarrow \left( {16 - x} \right) \times \left( {x + 1} \right) = 15 \times x\]
Now, use property of algebra $\left( {a + b} \right) \times \left( {c + d} \right) = a\left( {c + d} \right) + b\left( {c + d} \right)$
\[
   \Rightarrow 16\left( {x + 1} \right) - x\left( {x + 1} \right) = 15x \\
   \Rightarrow 16x + 16 - {x^2} - x = 15x \\
   \Rightarrow - {x^2} + 16 + 15x = 15x \\
 \]
Subtract 15x from both sides of equation,
\[
   \Rightarrow - {x^2} + 16 + 15x - 15x = 15x - 15x \\
   \Rightarrow - {x^2} + 16 = 0 \\
\]
If we shift any number or variables from one side of the equation to the other side, the sign becomes changed.
Shift 16 from one side to other side of equation,
\[ \Rightarrow - {x^2} = - 16\]
Negative sign cancel from both sides,
\[ \Rightarrow {x^2} = 16\]
Take a square root,
\[
   \Rightarrow x = \sqrt {16} \\
   \Rightarrow x = \pm 4 \\
\]
We know the square root of any positive number gives both positive and negative values.
We can also check the both values of x by putting into the equation.
Now, both values are correct.
So, the values of x are 4 or -4.

Note-In such types of problems we should remember some points. First, if we shift the variables and numbers from one side of the equation to the other side, the sign of variables and numbers become changed. Second, if we take a square root of any positive number give the both values positive and negative.