4x + 3( x - 1/5) ≥ 1/5( 2 + x )
Just simplify it
4x + 3x - 3/5 ≥ 2/5 + x/5
don't like those fractions? , multiply each term by 5
20x + 15x - 3 ≥ 2 + x
I am sure you can handle it from there, let me know what answer you get
What is the minimum value of x for this inequality?
4x + 3( x - 1/5) ≥ 1/5( 2 + x )
How in the hell do I do this? Couldn't there be an infinite amount of numbers??
4 answers
I don't understand, what do you mean?
7x - 3/5 ≥ 2/5 + x/5
ok, let's go with your messy-looking equation
7x - 3/5 ≥ 2/5 + x/5
7x - x/5 ≥ 2/5 + 3/5
(35/5)x - (1/5)x ≥ 5/5
(34/5)x ≥ 1
now multiply each side by 5
34x ≥ 5
now divide each side by 34
x ≥ 5/34
starting from my simplified version of
20x + 15x - 3 ≥ 2 + x
20x + 15x - x ≥ 5
34x ≥ 5
x ≥ 5/34 , same result, less messy calculations
7x - 3/5 ≥ 2/5 + x/5
7x - x/5 ≥ 2/5 + 3/5
(35/5)x - (1/5)x ≥ 5/5
(34/5)x ≥ 1
now multiply each side by 5
34x ≥ 5
now divide each side by 34
x ≥ 5/34
starting from my simplified version of
20x + 15x - 3 ≥ 2 + x
20x + 15x - x ≥ 5
34x ≥ 5
x ≥ 5/34 , same result, less messy calculations
1 answer