Question

Solve the following equations,
$
  y + \sqrt {{x^2} - 1} = 2 \\
  \sqrt {x + 1} - \sqrt {x - 1} = \sqrt {y.} \\
 $
\[
  {\text{(A) }}x = 2;y = 5 \\
  {\text{(B) }}x = 1;y = 2 \\
  {\text{(C) }}x = \dfrac{5}{3};\dfrac{2}{3} \\
  {\text{(D) }}x = 3;y = 3 \\
 \]

Answer 1

Hint: In the current form, equations are quadratic form if expanded. These two variable type equations can be reduced to linear form by substitution. After substitution, we will get two equations with two variables, which are easy to solve.

Complete step-by-step answer:
Firstly, rewrite the equations given in the question as,
$
  y + \sqrt {{x^2} - 1} = 2{\text{ }}...............{\text{ (A)}} \\
  \sqrt {x + 1} - \sqrt {x - 1} = \sqrt y {\text{ }}........{\text{(B)}} \\
 $
Now, these can be written as,
$
  y + \sqrt {(x + 1)(x - 1)} = 2{\text{ }}.................{\text{(1)}} \\
  \sqrt {x + 1} - \sqrt {x - 1} = \sqrt {y.} {\text{ }}.................{\text{(2)}} \\
 $
In equation (1), we used the identity ${p^2} - {q^2} = (p + q)(p - q)$to simplify. Now, assuming that $(x + 1) = a$ and $(x - 1) = b$, then the equation (1) and (2) will become as,
$
  y + \sqrt {ab} = 2{\text{ }}.................{\text{(3)}} \\
  \sqrt a - \sqrt b = \sqrt {y.} {\text{ }}...........{\text{(4)}} \\
 $
Also, we now know that,$a + b = 2x{\text{ and }}ab = {x^2} - 1$
Now, squaring the equation (4) on both sides, we get {Using the identity ${(p - q)^2} = {p^2} - 2pq + {q^2}$}
$
   \Rightarrow \sqrt a - \sqrt b = \sqrt y \\
   \Rightarrow a + b - 2\sqrt {ab} = y{\text{ }}.............{\text{ (5)}} \\
 $
From equation (3), we know that $\sqrt {ab} = 2 - y$, so putting this value of $\sqrt {ab} $ in equation (5), we get,
$
   \Rightarrow a + b - 2 = 2 - y \\
   \Rightarrow a + b = 4 - y{\text{ }}............{\text{ (6)}} \\
 $
Putting the value of $a + b = 2x$, the equation (6) will be,
\[
   \Rightarrow 2x = 4 - y \\
   \Rightarrow y = 4 - 2x{\text{ }}......................{\text{(7)}} \\
 \]
Now putting this value in equation (A), we will get,
\[
   \Rightarrow 4 - 2x + \sqrt {{x^2} - 1} = 2 \\
   \Rightarrow \sqrt {{x^2} - 1} = 2x - 2 \\
 \]
Now, rearranging and squaring both the sides, we will get
\[
   \Rightarrow \sqrt {{x^2} - 1} = 2(x - 1) \\
   \Rightarrow {x^2} - 1 = 4{(x - 1)^2} \\
 \]
Expanding {Using the identity ${(p - q)^2} = {p^2} - 2pq + {q^2}$} and rearranging the terms $x$ and ${x^2}$, we will get
\[
   \Rightarrow {x^2} - 1 = 4({x^2} - 2x + 1) \\
   \Rightarrow 3{x^2} - 8x + 5 = 0 \\
 \]
Factorising the above expression to get the following expression,
\[ \Rightarrow 3{x^2} - 3x - 5x + 5 = 0\]
Now take the common factors and resolve them. We get
\[
   \Rightarrow 3x(x - 1) - 5(x - 1) = 0 \\
   \Rightarrow (3x - 5)(x - 1) = 0 \\
 \]
This will give two values of $x$ as,
\[x = 1,\dfrac{5}{3}\]
Now putting these values of $x$in equation (7), we the value of $y$ as,
$
  {\text{when, }}x = 1,y = 2 \\
  {\text{and when }}x = \dfrac{5}{3},y = 4 - \dfrac{{10}}{3} = \dfrac{2}{3} \\
 $
This way,
$
  x = 1,\dfrac{5}{3} \\
  y = 2,\dfrac{2}{3} \\
 $
Therefore correct options are (B) and (C).

Note: In case of two variable equations, solving these types of problems, one should remember the initial equations, so that it would be easy to replace the one variable in terms of the other. The assumption of other variables apart from the variables in the question will minimise the order of the equation. In the above problem, we have reduced order of the equations from second to one. And due to this, the quadratic equation has become a linear one, which is easy to solve. Reading the problem carefully solves half of the problem.