How do you solve #3(6- x)+ 2x = 15 #?

distribute the bracket.

#rArr18-3x+2x=15#

collecting terms in x on the left side and numeric values on the right side.

#rArr18-x=15#

subtract 18 from both sides.

#cancel(18)cancel(-18)-x=15-18#

#rArr-x=-3rArrx=3#

#color(blue)"As a check"#

substitute this value into the left side and if it equals the right side then it is the solution.

#3(6-3)+(2xx3)=(3xx3)+6=9+6=15#

#rArrx=3" is the solution"#

To solve this problem, you have to get rid of the parentheses first. To do this, you have to distribute the 3 to all the terms in the parentheses. After you have done this, you will get #18-3x+2x=15#.

Next, you have to check if there are any like terms in the equation. Since there are, #3x and 2x#, you can combine those two. After this step, you will get #18-x=15#.

To solve a problem, you have to get the variable by itself. To do this in this problem, you have to get rid of the 18. To get rid of the 18, you have to subtract it from the left side. But remember, whatever you do to one side, you have to do to the other, too. So subtract 18 from the right side, too. You will have #-x=-3#.

You are just one step away from the answer. You cannot have the variable as a negative number. You have to make it positive. To do this, you have to divide both the left and the right sides by -1. This gets rid of the negative sign next to the variable and makes it positive.

After you have completed all of these steps, you have found your answer, which is #x=3#.