# How do you solve 3-x=18-4x?

Oct 28, 2015

x=5
Method shown in 'over the top' detail so that you can see what is happening. You should then be able to solve others more easily.

#### Explanation:

The brackets are used to show grouping to help understand what is happening.

Note: add is the same as 'put with'
subtract is the same as 'remove'

To get of something one side of the = sign you do the opposite action. Also, what you do to one side of the = sign you also do to the other. So if something is add (put with) on one side then you have to remove it (subtract) to get rid of it. That action applied to both sides of the = sign effectively moves it from one side to the other. It if it multiply then you have to divide and so on.

$\left(3 - x\right) = \left(18 - 4 x\right)$........................ (1)

To get 'rid of' the$\text{ }$ 'remove' $\text{ } 4 x$ ( which is$- 4 x$) on the right hand side we put it back, which is $+ 4 x$

Add ($+ 4 x$) to both sides giving:
$\left(3 - x\right) + 4 x = \left(18 - 4 x\right) + 4 x$........................(2)

$3 + \left(- x + 4 x\right) = 18 + \left(- 4 x + 4 x\right)$.............................(${2}_{\text{regrouped}}$)

but:

$- x + 4 x = 3 x$ ..........................(3)
$- 4 x + 4 x = 0$...........................(4)

Substituting (3) and (4) into (${2}_{\text{regrouped}}$) giving:

$3 + 3 x = 18 + 0 \text{ }$
which is the same as $\text{ } 3 + 3 x = 18$......(5)

Subtract 3 from both sides.
Grouping again as:
$\left(3 + 3 x\right) - 3 = \left(18\right) - 3$

giving:

$3 x = 15$ ........................(6)

But $3 x$ means that $x$ is multiplied by 3. To get rid of the three we do the revers of multiply, which is divide by 3. Which is also multiply by$\frac{1}{3}$.

Multiply both sides of (6) by $\frac{1}{3}$ giving;

$\frac{1}{3} \times \left(3 x\right) = \frac{1}{3} \times \left(15\right)$

$\frac{3}{3} \times x = \frac{15}{3}$

$1 \times x = 5$

$x = 5$