Question
Solve the inequality and represent the solution graphically on number line: 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
✓
Answer
We have 2(x - 1) < x + 5, and 3(x + 2) > 2 - x
$ \Rightarrow 2x - 2 < x + 5$ and 3x + 6 > 2 - x
$ \Rightarrow x < 7$ and 4x > -4
$ \Rightarrow x < 7$ and x > -1
$ \Rightarrow 2x - 2 < x + 5$ and 3x + 6 > 2 - x
$ \Rightarrow x < 7$ and 4x > -4
$ \Rightarrow x < 7$ and x > -1
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the mean deviation about the mean for the following data:
6, 7, 10, 12, 13, 4, 8, 12
6, 7, 10, 12, 13, 4, 8, 12
Differentiate the following from first principle:
→$\frac{1}{\sqrt{\text{x}}}$
Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
Evaluate the following:
→$\sum_\limits{\text{n}=2}^{10}(4^\text{n})$
Find the middle term in the expansion of:
$\Big(\frac{\text{a}}{\text{x}}+\text{bx}\Big)^{12}$
Sketch the graphs of the following trigonometric functions:
$\psi\text{(x)}=\cos3\text{x}$
→$\psi\text{(x)}=\cos3\text{x}$
Find the equation of the circle which circumscribes the triangle formed by the lines
2x + y - 3 = 0, x + y - 1 = 0 and 3x + 2y - 5 = 0
2x + y - 3 = 0, x + y - 1 = 0 and 3x + 2y - 5 = 0
In what ratio is the line joining A (-1, 1) and B(5, 7) divided by the line x + y = 4?
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b}$ prove that
$\cos(\alpha-\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
→$\cos(\alpha-\beta)=\frac{\text{a}^2+\text{b}^2-2}{2}$
If the real part of $\frac{\bar{\text{z}}+2}{\bar{\text{z}}-1}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.
→