Question

$2 x - 3 \leqslant x + 1 \leqslant 4x + 7$
linear inequations​

Answer 1

Given :

$2 x - 3 \leqslant x + 1 \leqslant 4x + 7$

$2 x - 3 \leqslant x + 1$

• Let's solve this one

$2 x - x\leqslant 3+ 1$

$x\leqslant 4$

• Now let's solve this one too

$x + 1 \leqslant 4x + 7$

$x-4x\leqslant 7-1$

$- 3x\leqslant 6$

$3x \geqslant -6$

$x\geqslant \dfrac{( - 6)}{3}$

$x \geqslant - 2$

Hence, x lies between -2 ≤ x ≤ 4

Points to remember :

Linear inequalities is same as linear equations, but instead of fundamental operation these are connected with inequalities (<,>,≤,≥). And also, if we multiply or divide by negative sign, inequality get reversed.

Answer 2

Step-by-step explanation:

$\bf\LARGE\underline{\underline{\red{Question :}}}$

• $\tt\large2 x - 3 \leqslant x + 1 \leqslant 4x + 7$ linear inequations.

$\bf\LARGE\underline{\underline{\green{Solution :}}}$

$\tt\large2 x - 3 \leqslant x + 1 \leqslant 4x + 7$

Let's solve the first part :

$\implies\large\tt 2x - 3 \: \leqslant \: x + 1$

$\implies\large\tt 2x - x \leqslant 3 + 1$

$\implies\large\tt x \leqslant 4$

Now, let's solve the second part :

$\implies\tt\large \: x + 1 \leqslant 4x + 7$

$\implies\large\tt x - 4x \leqslant 7 - 1$

$\implies\large\tt -3x \leqslant 6$

$\implies\large\tt \: x \geqslant \frac{6}{ - 3}$

$\implies\large\tt x \geqslant -2$

Therefore, x lies between $\large\tt -2 \leqslant x \leqslant 4$