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Question

Question: Solve for \(x\) in \(2x + 3y = 6\)....

Solve for xx in 2x+3y=62x + 3y = 6.

Explanation

Solution

The above given expression is an example of a two step equation. In order to solve it we need to manipulate the given equation in such a way that we should get xx by itself. In order to get xx by itself we can perform any arithmetic operations on both LHS and RHS equally at the same time such that the equality of the given equation doesn’t change.

Complete step by step solution:
Given
2x+3y=6....................................(i)2x + 3y = 6....................................\left( i \right)
Now in order to solve the given equation we need to solve for xx.
Such that we have to manipulate the given equation in terms of only xx, which can be achieved by performing different arithmetic operations on both LHS and RHS equally.
So to isolate the xxterm from equation (i) first we have to subtract the term 3y3y from both LHS and RHS.
Such that:

2x+3y3y=63y 2x=63y.......................................(ii) \Rightarrow 2x + 3y - 3y = 6 - 3y \\\ \Rightarrow 2x = 6 - 3y.......................................\left( {ii} \right) \\\

Now to isolate the xx term from equation (ii) we have to divide 2 from both the LHS and RHS.
Such that:
2x=63y 2x2=63y2 x=623y2........................(iii)  \Rightarrow 2x = 6 - 3y \\\ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{6 - 3y}}{2} \\\ \Rightarrow x = \dfrac{6}{2} - \dfrac{{3y}}{2}........................\left( {iii} \right) \\\
Now on simplifying the RHS we can write:
x=33y2........................(iv)\Rightarrow x = 3 - \dfrac{{3y}}{2}........................\left( {iv} \right)
Therefore on solving 2x+3y=62x + 3y = 6 we get x=33y2x = 3 - \dfrac{{3y}}{2}.

Note:
A two-step equation is an algebraic equation which can be solved in two steps. The equation is said to be true when we find the value of the variable which makes the equation true.
We can also check if the value of the variable that we got is true or not by substituting the value of the variable back into the equation and checking whether it satisfies the given equation or not.