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Question

Question: How do you solve \(5x - 3 = 13 - 3x?\)...

How do you solve 5x3=133x?5x - 3 = 13 - 3x?

Explanation

Solution

Eliminate the term which is the coefficient of “x” in the right hand side (R.H.S) and also eliminate the constant from left hand side (L.H.S) by doing addition with their respective additive inverse to the both sides. Finally you will get an equation in which the left hand side will consist only of the term which is the coefficient of “x” and the right hand side will consist of only constant, then divide both sides with the coefficient of “x”.

Complete step by step solution:
In order to solve 5x3=133x,5x - 3 = 13 - 3x, we need to go through some steps to eliminate the term which is the coefficient of “x” in the right hand side and the constant from the left hand side. We will start by eliminating 3 - 3 in the L.H.S. of the given equation by adding 33 to both sides.

We are adding 33 because it is the additive inverse of 3 - 3 and adding to both sides because only then it will not affect the real equation and maintain the respective balance of the equation.

First rearranging the equation and then adding 33 both sides, we will get
$
\Rightarrow 5x - 3 = 13 - 3x \\

\Rightarrow 5x - 3 = - 3x + 13 \\

\Rightarrow 5x - 3 + 3 = - 3x + 13 + 3 \\
Solvingfurther, Solving further,
\Rightarrow 5x - 0 = - 3x + 16 \\

\Rightarrow 5x = - 3x + 16 \\
Now,adding Now, adding3xtobothsidesinordertoeliminateto both sides in order to eliminate - 3x,wewillget, we will get
\Rightarrow 5x = - 3x + 16 \\

\Rightarrow 5x + 3x = - 3x + 16 + 3x \\

\Rightarrow 5x + 3x = - 3x + 3x + 16 \\

\Rightarrow 8x = 0 + 16 \\

\Rightarrow 8x = 16 \\
Nowdividingbothsideswith Now dividing both sides with8$, which is the coefficient of “x”, we will get

8x=16 8x8=168 x=2 \Rightarrow 8x = 16 \\\ \Rightarrow \dfrac{{8x}}{8} = \dfrac{{16}}{8} \\\ \Rightarrow x = 2 \\\

\therefore we get the required solution for the equation 5x3=133x5x - 3 = 13 - 3x which is x=2x = 2

Note: If bb is the additive inverse of aa , then the sum of a  and  ba\;{\text{and}}\;b will be equals to 00 a+b=0 \Rightarrow a + b = 0 then the value bb will be given as b=0a=a \Rightarrow b = 0 - a = - a. There also exists multiplicative inverses, if aa is the number then its multiplicative inverse (say bb) will be given as b=1ab = \dfrac{1}{a}