Solution: $\left|\begin{array}{ccc} a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0 \end{array}\right|$ $\begin{array}{l} \left.=\left(\frac{1}{x...

### Solve |x| > 4, when x Ο΅ R.

Answer : |x| > 4 Square β x2 > 16 β x2 β 16 > 0 β x2 β 42 > 0 β (x + 4)(x β 4) > 0 Observe that when x is greater than 4, (x + 4)(x β 4) is positive And for each root the sign changes...

### Solve |x| < 4, when x Ο΅ R.

Answer : |x| < 4 Square β x2 < 16 β x2 β 16 < 0 β x2 β 42 < 0 β (x + 4)(x β 4) < 0 Observe that when x is greater than 4, (x + 4)(x β 4) is positive And for each root the sign changes...

### Solve x/x-5>1/2, when x Ο΅ R.

Answer :Observe thatΒ Β Β Β Β Β is zero at x = -5 and not defined at x = 5 Hence plotting these two points on number line Now for x > 5Β is positive For every root and not defined value of the sign...

### Solve 3/x-2< 1, when x Ο΅ R.

Answer :Observe that zero at x = 5 and not defined at x = 2 Hence plotting these two points on number line Now for x > 5,Β is negative For every root and not defined value of the sign will change...

### Solve x + 5 > 4x β 10, when x Ο΅ R.

Answer : x + 5 > 4x β 10 β 5 + 10 > 4x β x β 15 > 3x Divide by 3 β 5 > x β x < 5 x < 5 means x is from -β to 5 that is x β (-β, 5) Hence solution of x + 5 > 4x β 10 is x β (-β,...

### Solve β4x > 16, when x Ο΅ Z.

Answer : We have to find integer values of x for which -4x > 16 (why only integer values because it is given that x β Z that is set of integers) -4x > 16 β -x > 4 β x < -4 The integers...

### Solve the system of in equation x β 2 β₯ 0, 2x β 5 β€ 3.

Answer : We have to find values of x for which both the equations hold true x β 2 β₯ 0 and 2x β 5 β€ 3 We will solve both the equations separately and then their intersection set will be solution of...

### Find the solution set of the in equation

### Find the solution set of the in equation

Answer :Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β means we have to find values of x for whichΒ Β Β Β Β Β Β Β is negative Observe that the numerator |x β 2| is always positive because of mod, hence for negative quantity the...

### Find the solution set of the in equation |x β 1| < 2.

Answer : |x β 1| < 2 Square both sides β (x β 1)2 < 4 β x2 β 2x + 1 < 4 β x2 β 2x β 3 < 0 β x2 β 3x + x β 3 < 0 β x(x β 3) + 1(x β 3) < 0 β (x + 1)(x β 3) < 0 Observe that when...

### Find the solution set of the in equation

Answer : We have to find values of x for which is less than zero that is negative Now for negative x β 2 should be negative that is x β 2 < 0 β x β 2 < 0 β x < 2 Hence x should be less than...