What is 4(x+3)-7>= x+3(x+1) ?

The solution to 4(x+3)-7>= x+3(x+1) is True for all x

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4(x+3)-7>= x+3(x+1)

4 (x + 3) - 7 \geq x + 3 (x + 1)

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

4 (x + 3) - 7 \geq x + 3 (x + 1)

Expand

3 (x + 1) : 3 x + 3

4 x + 5 \geq x + 3 x + 3

Add similar elements:

x + 3 x = 4 x

4 x + 5 \geq 4 x + 3

Expand

4 (x + 3) - 7 : 4 x + 5

4 x + 5 \geq 4 x + 3

Move

5

to the right side

4 x \geq 4 x - 2

Move

4 x

to the left side

0 \geq - 2

Therefore, the solution is

True for all x

- 4 x = \frac{11}{5}

\frac{x}{4} + 1 < 1

8 y^{3} + 1

30 \cdot 4

21 x - 17 > 25

What is 4(x+3)-7>= x+3(x+1) ?
The solution to 4(x+3)-7>= x+3(x+1) is True for all x