Question

How do you solve for \[y\] in \[2x + 3y = 4\]?

Answer 1

Hint: Here we find the required value by isolating the variable that needs to be found out. For that, we will take all the terms on the right-hand side except the term having \[y\] variable. Then we will remove the coefficient of \[y\] variable and bring it in the denominator of the other side to get the required answer.

Complete step-by-step answer:
We have to solve the equation \[2x + 3y = 4\] for \[y\].
First, we will subtract \[2x\] from both sides. Therefore, we get
\[3y = 4 - 2x\]
Now, we will take the coefficient of \[y\] in the denominator of the right-hand side.
Dividing both sides by 3, we get
\[ \Rightarrow y = \dfrac{{4 - 2x}}{3}\]
So, we get the solution as\[y = \dfrac{{4 - 2x}}{3}\].
By substituting different values of \[x\] we get different values for \[y\].
 For example if we substitute \[x = 0\] in the equation \[y = \dfrac{{4 - 2x}}{3}\], we get
\[y = \dfrac{{4 - 2\left( 0 \right)}}{3}\]
\[ \Rightarrow y = \dfrac{4}{3}\]
So, when \[x = 0\] then \[y = \dfrac{4}{3}\].
Now if we substitute \[x = 1\] in the equation \[y = \dfrac{{4 - 2x}}{3}\], we get
\[y = \dfrac{{4 - 2\left( 1 \right)}}{3} = \dfrac{{4 - 2}}{3}\]
\[ \Rightarrow y = \dfrac{2}{3}\]

So, when \[x = 1\] then \[y = \dfrac{2}{3}\].

Therefore, we can find an infinite value of \[y\] by using the same technique.

Note:
The equation that can be written in a form of \[ax + by + c = 0\] is known as Linear equation in two variables where,\[a,b,c\] are real numbers and \[a,b\] are not both zero and also the coefficient of \[x\] and \[y\] variables. To find the solution of the Linear equation in two variables we use the elimination method. Linear equations are used to convert a statement into an equation so that it can be solved. The solution of a Linear equation in two variables is always in pairs where one value is for \[x\] and another for\[y\]. The equation is Linear because the highest power of the variables is 1. If we form a graph of the Linear equation we get a straight line.