Question

How do you solve $4\left( {k + 2} \right) = 3$?

Answer 1

Hint: First step is to apply distributivity property in the given equation, i.e., multiplying each addend individually by the number and then adding the products together. Next, we have to isolate the variable terms on one side by performing the same mathematical operations on both sides of the equation. Next step is to isolate the constant terms on the other side by performing the same mathematical operations on both sides of the equation. Next step is to make the coefficient of the variable equal to $1$ using multiplication or division property.

Complete step by step answer:
Given an algebraic equation is $4\left( {k + 2} \right) = 3$.
We have to find the value of $k$.
First, we have to apply the distributive property in the given equation, i.e., multiplying each addend individually by the number and then adding the products together.
$4\left( k \right) + 4\left( 2 \right) = 3$
Calculate the product of each term in the left hand side of the equation.
$ \Rightarrow 4k + 8 = 3$
Now we have to isolate the constant terms on the other side by performing the same mathematical operations on both sides of the equation.
So, subtracting $8$ from both sides of the equation, we get
$ \Rightarrow 4k + 8 - 8 = 3 - 8$
It can be written as
$ \Rightarrow 4k = - 5$
Now we have to make the coefficient of the variable equal to $1$ using multiplication or division property.
So, dividing both sides of the equation by $4$, we get
\[ \Rightarrow \dfrac{{4k}}{4} = \dfrac{{ - 5}}{4}\]
$\therefore k = - \dfrac{5}{4}$

Therefore, $k = - \dfrac{5}{4}$ is the solution of $4\left( {k + 2} \right) = 3$.

Note: An algebraic equation is an equation involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS).
In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation.