How do you solve #4(k+2)=3#?

1 Answer
Jan 19, 2017

Answer:

#k=-5/4#

Explanation:

distribute the bracket.

#rArr4k+8=3#

subtract 8 from both sides of the equation.

#4xcancel(+8)cancel(-8)=3-8#

#rArr4x=-5#

To solve for x, divide both sides by 4

#(cancel(4) x)/cancel(4)=(-5)/4#

#rArrx=-5/4#

#color(blue)"As a check"#

Substitute this value for x into the left side of the equation and if it is equal to the right side then it is a solution.

#x=-5/4to4(-5/4+2)=-5+8=3=" right side"#

#rArrx=-5/4" is the solution"#