Question

Write any two solutions of \[2x + 3y = 6\]

Answer 1

Hint: We find the value of one of the variables in terms of another variable. Substitute values for one of the variables and calculate the corresponding values for other variables. Write the solutions in terms of values of x and y.
Solution of any equation means those values of the given variables that satisfy the equation. On substitution of the corresponding values of variables the equation equals to zero.

Complete step-by-step answer:
We are given the equation \[2x + 3y = 6\]
Shift the value ‘2x’ to RHS of the equation
\[ \Rightarrow 3y = 6 - 2x\]
Divide both sides by 3
\[ \Rightarrow \dfrac{{3y}}{3} = \dfrac{{6 - 2x}}{3}\]
Cancel the same terms from numerator and denominator from both sides of the equation.
\[ \Rightarrow y = \dfrac{{6 - 2x}}{3}\] ………...… (1)
FIRST SOLUTION:
We have the value of y in terms of x as\[y = \dfrac{{6 - 2x}}{3}\]
Substitute the value of \[x = 1\]
\[ \Rightarrow y = \dfrac{{6 - 2 \times 1}}{3}\]
Calculate the product in numerator
\[ \Rightarrow y = \dfrac{{6 - 2}}{3}\]
Calculate the difference in numerator
\[ \Rightarrow y = \dfrac{4}{3}\]
So, the solution is \[x = 1,y = \dfrac{4}{3}\] ……...… (2)
SECOND SOLUTION:
We have the value of y in terms of x as\[y = \dfrac{{6 - 2x}}{3}\]
Substitute the value of \[x = 2\]
\[ \Rightarrow y = \dfrac{{6 - 2 \times 2}}{3}\]
Calculate the product in numerator
\[ \Rightarrow y = \dfrac{{6 - 4}}{3}\]
Calculate the difference in numerator
\[ \Rightarrow y = \dfrac{1}{3}\]
So, the solution is \[x = 2,y = \dfrac{1}{3}\] …….… (3)

\[\therefore \]From equations (2) and (3) we can write two solutions of the equation \[2x + 3y =6\] are \[x = 1,y = \dfrac{4}{3}\] and \[x = 2,y = \dfrac{1}{3}\]

Note: Students make mistakes while shifting the values from one side of the equation to another side as they don’t change signs from negative to positive and vice versa. Also, while cross multiplying always brings the value of the numerator to the denominator and vice versa.
Students can make any number of solutions by substituting any value of one of the variables and find the corresponding value of another variable.