First slide

Introduction to linear inequalities

Question

The system of equation $| x - 1 | + 3 y = 4, x - | y - 1 | = 2$ has

Moderate

A

no solution
B

a unique solution
C

two solutions
D

more than two solutions

Solution

The given equation is $| x - 1 | + 3 y = 4$
$\Rightarrow$ ${\begin{matrix} x + 3 y = 5, x \geq 1 ....... (i) \\ - x + 3 y = 3, x < 1 ......... (i i) \end{matrix}$

and    $x - | y - 1 | = 2$

$\Rightarrow$ ${\begin{matrix} x - y = 1, y \geq 1 .......... (i i i) \\ x + y = 3, y < 1 ........... (i v) \end{matrix}$

Solving Eqs. (i) and (iii), we get

                $x = 2, y = 1$

Solving Eqs. (i) and (iv), we get

    $x = 2, y = 1$ no solution       $(∵ x \geq 1, y < 1)$

Solving Eqs. (ii) and (iv), we get

$x = \frac{5}{2}, y = \frac{3}{2}$ no solution           $(∵ x < 1, y < 1)$

Hence, solution is $x = 2, y = 1$   (a unique solution).

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