How do you solve #6j + 8- 11= 0#?

1 Answer
Ryuu
Apr 6, 2017

Add like terms, do BEDMAS backwards on the equaling term, and remove coefficient. In this case, #j=1/2#.

Explanation:

Solving an equation with it being equaled to something implies determining the value of the variable.

We can solve #6j + 8 - 11 = 0# by isolating #j#. This can be accomplished performing BEDMAS/PEMDAS backwards on the equaling value.

Before we do that, we have to add like terms. So let's bring all the similar terms on one side of the equal sign, and the other terms on the other side. When this happens, the signs are inverted.

#6j = 11-8#

#6j = 3#

Now let's divide #3# by #6# to remove the coefficient.

#(6j)/6 = 3/6#

#j = 3/6#

We can simplify #3/6#, so let's do that.

#j = 1/2#

And there we go: #6=j#. We can double-check our work by subbing in #6# as #j# in the original equation.

#6j + 8 - 11 = 0#

#6(1/2) + 8 - 11 = 0#

#3 + 8 - 11 = 0#

#11 - 11 = 0#

#0=0#

Therefore, we can conclude that #j=1/2#.

Hope this helps :)

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