x = 11
Thrice A number divided by five is three
2n-4
2x-5=25 2x = 30 x=15
are you asking for solutions to 5x + 5 = 2y? (x,y) = (1,5), (3,10), (5,15), etc.
-9
If the number is 'z', then five less than twice the number is ( 2z - 5 ).
10
x = 11
Thrice A number divided by five is three
blah
2n-4
Any number that's −30 or less would fit.
-5
2x-5=25 2x = 30 x=15
are you asking for solutions to 5x + 5 = 2y? (x,y) = (1,5), (3,10), (5,15), etc.
To write out the given statement algebraically, you would start by defining the variable for the number, let's say it is represented by 'x'. The equation would be: 6x - 5 < 2x + 10. This equation represents the statement "Six times a number minus five is less than twice the number plus ten" in algebraic form. To solve this inequality, you would isolate the variable 'x' by performing operations to simplify and find the range of values that satisfy the inequality.