Question 1
18924

Solve the inequality \(\dfrac{3}{{x + 1}} \le 1\).

\(x < - 1\) or \(x \ge 2\)  

Question 2
18925

Solve the inequality \(\dfrac{{4x}}{{x - 2}} \ge 1\).

\(x \le - \dfrac{2}{3}\) or \(x > 2\)

Question 3
18926

Solve the inequality \(\dfrac{1}{x} > \dfrac{1}{{x - 1}}\).

\(0 < x < 1\)

Question 4
18927

Solve the inequality \(\dfrac{{2 - {x^2}}}{x} > 1\).

\(x < - 2\), \(0 < x < 1\)

Question 5
13856

The solutions to \(\dfrac{x}{{x + 1}} > 0\) are? 

\( x < -1\) or \(x > 0\) 

Question 6
13857

The solutions to \(\dfrac{{x - 3}}{{x + 2}} > 1\) are? 

\(x < - 2\)

Question 7
13858

The solutions to \(\dfrac{3}{{1 - x}} \le 2\) are? 

\(x \le - \frac{1}{2}\) or \(x > 1\) 

Question 8
13859

The solutions to \(\dfrac{{{x^2}}}{{{x^2} - 1}} \le 1\) are? 

\( - 1 < x < 1\)

Question 9
13860

The solutions to \(\dfrac{{x - 1}}{{x + 1}} + 1 < 0\) are? 

\( - 1 < x < 0\)

Question 10
122934

(i) On a number plane, sketch the graphs of \(y=5-2 x\) and \(y=\dfrac{2}{x}\).

(ii) Using part (i) or otherwise write down the values of \(x\) for which \(5-2 x<\dfrac{2}{x}\)

i) Refer to worked solution

ii) \(0<x<\dfrac{1}{2} \text { and } x>2\)

Question 11
122935

(i) On a number plane, sketch the graphs of \(y=x+2\) and \(y=\dfrac{2 x}{x-1}\).

(ii) Using part (i) or otherwise write down the values of \(x\) for which \(x+2>\dfrac{2 x}{x-1}\)

\(-1<x<1 \text { and } x>2\)

Question 12
122936

Solve the inequality \(\dfrac{2-x}{x} \geq 1\)

\(0<x \leq 1 \quad x \neq 0\)

Question 13
122937

Solve the inequality \(\dfrac{x+1}{x-2} \leq 2\)

\(x<2 \text { or } x \geq 5 \quad x \neq 2\)

Question 14
122938

(i) On a number plane sketch the graphs of \(y=1-x\) and \(y=\dfrac{1-3 x}{1+3 x}\).

(ii) Using part (i) or otherwise write down the values of \(x\) for which \(1-x>\dfrac{1-3 x}{1+3 x}\)

i) Refer to worked solution

ii) \(x<-\dfrac{1}{3} \text { and } 0<x<\dfrac{5}{3}\)

Question 15
122939

Solve the inequality \(\dfrac{1}{x+1}+\dfrac{1}{x}<\dfrac{1}{x(x+1)}\)

\(x<-1\)

Question 16
122940

Solve the inequality \(\dfrac{x-1}{x+2} \leq \dfrac{x-2}{x+1}\)

\(-2<x<-1\)