What is -4m+21(m-1)<= 15m+2(m-1) ?

The solution to -4m+21(m-1)<= 15m+2(m-1) is True for all m

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-4m+21(m-1)<= 15m+2(m-1)

- 4 m + 21 (m - 1) \leq 15 m + 2 (m - 1)

True for all m

Interval Notation

(- \infty, \infty)

Solution steps

- 4 m + 21 (m - 1) \leq 15 m + 2 (m - 1)

Expand

21 (m - 1) : 21 m - 21

- 4 m + 21 m - 21 \leq 15 m + 2 (m - 1)

Add similar elements:

- 4 m + 21 m = 17 m

17 m - 21 \leq 15 m + 2 (m - 1)

Expand

2 (m - 1) : 2 m - 2

17 m - 21 \leq 15 m + 2 m - 2

Add similar elements:

15 m + 2 m = 17 m

17 m - 21 \leq 17 m - 2

Move

21

to the right side

17 m \leq 17 m + 19

Move

17 m

to the left side

0 \leq 19

Therefore, the solution is

True for all m

0 \geq \frac{5 - x}{2} \geq - 10

\frac{( n + 2 ) !}{( n + 1 ) !}

(- 1)^{\frac{4}{5}}

32 e^{2} - 50

36 x^{2} - y^{6}

What is -4m+21(m-1)<= 15m+2(m-1) ?
The solution to -4m+21(m-1)<= 15m+2(m-1) is True for all m