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To solve two step linear inequalities, we have to do the addition or subtraction first using opposite operations, and then use the multiplication or division according to the given equation.The opposite operation of addition is subtraction and vice versa.

Similarly, the opposite operation of multiplication is division and vice versa.

Transpose +7 to other side, we get,

6x > 4x + 3 -7

6x > 4x - 4

Now transpose 4x to left side so that like terms on one side and constants on other side

6x - 4x > -4

2x > -4 divide both side by 2

$\frac{2x}{2} > \frac{-4}{2}$

∴ x > -2

First, open the parenthesis

x + 5 - 7x + 14 $\geq$ 4x + 9

Add the like terms

-6x + 19 $\geq$ 4x + 9

Transpose +19 to other side

-6x $\geq$ 4x + 9 -19

Now transpose 4x to left side so that variable terms on one side and constants on other side

-6x - 4x $\geq$ -10

-10x $\geq$ -10

Divide both side by -10. Dividing by negative number

$\frac{-10x}{-10} \leq \frac{-10}{-10}$

∴ x $\leq$ 1

$\frac{3x -4}{2}\geq \frac{x + 1}{4} - \frac{1}{1}$ ( just 1 means one over one)

First we will simplify the right side. There are subtraction of two fractions so find the LCD of 4 and 1, it will be 4

$\frac{3x -4}{2}\geq \frac{x + 1}{4} - \frac{4}{4}$

$\frac{3x -4}{2}\geq \frac{x + 1 - 4}{4}$

$\frac{3x -4}{2}\geq \frac{x - 3 }{4} $

Since we have fractions on both side , we will cross multiply

4(3x - 4) $\geq$ 2(x -3 )

12x - 16 $\geq$ 2x - 6

12x - 2x $\geq$ - 6 + 16

10x $\geq$ 10

$\frac{10x}{10} \geq \frac{10}{10}$

∴ x $\geq $ 1

On the left side we have subtraction of two fractions so find the LCD of 4 and 5, it will be 20

$\frac{5(x + 2) - 4x }{20} \geq 3$

$\frac{5x + 10 - 4x }{20} \geq 3 $

$\frac{x + 10 }{20} \geq 3 $

Now multiply both side by 20

(x + 10 ) $\geq$ 60

∴ x $\geq$ 50

From two step linear inequalities to Home

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