How do you solve #4(u+1)+5=6(u-1)+u#?

1 Answer
Apr 26, 2017

Answer:

#u=5#

Explanation:

Distribute brackets on both sides of the equation.

#4u+4+5=6u-6+u#

#"simplify both sides"#

#4u+9=7u-6#

#"subtract 7u from both sides"#

#4u-7u+9=cancel(7u)cancel(-7u)-6#

#rArr-3u+9=-6#

#"subtract 9 from both sides"#

#-3ucancel(+9)cancel(-9)=-6-9#

#rArr-3u=-15#

#"divide both sides by - 3"#

#(cancel(-9) u)/cancel(-9)=(-15)/(-3)#

#rArru=5#

#color(blue)"As a check"#

Substitute this value into the equation and if both sides are equal then it is the solution.

#"left side "=4(5+1)+5=(4xx6)+5=24+5=29#

#"right side "=6(5-1)+5=(6xx4)+5=29#

#rArru=5" is the solution"#