How do you solve 3x+16=5x+8?

Nov 8, 2015

More detail of process

$x = 4$

Explanation:

You need to end up with one $x$ on the left and everything else on the right (with no $x ' s$ on the right).

The existing LHS (left hand side) needs to have the 16 removed so that you just have the $x$ term.

Remove all of the 16 from LHS by subtraction. (Subtraction in this context is removal). Note that using subtraction in this context turns the existing value to 0 and anything + 0 does not change the overall value.

$\left(3 x + 16\right) - 16 = \left(5 x + 8\right) - 16$

What you do to the left you also do to the right to maintain the 'truth' of the equation. The brackets are there just to show you the original parts of the equation.

$3 x = 5 x - 8$

Now we need to 'remove' all of the $5 x$ from the right. Again we do so by subtraction.

$\left(3 x\right) - 5 x = \left(5 x - 8\right) - 5 x$

$- 2 x = - 8$

To end up with just $x$ on the left we don't remove all of it so subtraction does not work. We wish to remove part of it. That is the 2. This time it is different. In this context removal means change the (-2) into 1 as anything multiplied by 1 does not have its value changed.

Divide both sides by (-2)

$\frac{- 2}{- 2} x = \frac{- 8}{- 2}$

Divide a negative by another negative changes the outcome to positive.

$1 x = 4$ But 1 of $x$ is $x$ so we have

$x = 4$