Answer
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Hint: Here we'll proceed by taking one among the equations from the pair and convert it into a 3rd equation. Then we'll substitute the newly formed equation into another equation from the given pair of equations, it'll give the worth of 1 variable. After that substitute the value in the third equation to get the value of the second variable. Thus, we will get the required values of the equation.
Complete step by step solution:
Linear pairs of equations are equations that can be expressed as \[ax + by + c = 0\] where a, b and c are real numbers and both a, b are non-zero.
In this question, two equations are-
$ \Rightarrow 2x + y = 8$...................….. (1)
$ \Rightarrow - 2x + 3y = 12$................….. (2)
Firstly, we will take equation (2) and convert it,
$ \Rightarrow - 2x + 3y = 12$
Add $2x$ on both sides of the equation,
$ \Rightarrow - 2x + 3y + 2x = 12 + 2x$
Simplify the term,
$ \Rightarrow 3y = 12 + 2x$
Divide both sides by 3,
$ \Rightarrow y = \dfrac{{12 + 2x}}{3}$...............….. (3)
Now we will put the value of y from equation (3) in equation (1),
$ \Rightarrow 2x + \dfrac{{12 + 2x}}{3} = 8$
Take LCM on the left side,
$ \Rightarrow \dfrac{{6x + 12 + 2x}}{3} = 8$
Cross multiply the terms and simplify the equation,
$ \Rightarrow 8x + 12 = 24$
Move constant on right side,
$ \Rightarrow 8x = 12$
Divide both sides by 8,
$ \Rightarrow x = \dfrac{{12}}{8}$
Cancel out the common factors,
$ \Rightarrow x = \dfrac{3}{2}$
Here we will substitute the value of x in equation (3) to get the value of y,
$ \Rightarrow y = \dfrac{{12 + 2 \times \dfrac{3}{2}}}{3}$
Cancel out the common factors,
$ \Rightarrow y = \dfrac{{12 + 3}}{3}$
Subtract the value in the numerator,
$ \Rightarrow y = \dfrac{{15}}{3}$
Divide numerator by denominator,
$ \Rightarrow y = 5$
Hence, the values of $x$ and $y$ are $\dfrac{3}{2}$ and $5$ respectively.
Note: Whenever we face such sorts of problems the key concept is to use various methods of variable evaluation either by elimination or by substitution method. These methods will help in getting the proper track to gauge these equations involving two variables and reach the proper solution.
Complete step by step solution:
Linear pairs of equations are equations that can be expressed as \[ax + by + c = 0\] where a, b and c are real numbers and both a, b are non-zero.
In this question, two equations are-
$ \Rightarrow 2x + y = 8$...................….. (1)
$ \Rightarrow - 2x + 3y = 12$................….. (2)
Firstly, we will take equation (2) and convert it,
$ \Rightarrow - 2x + 3y = 12$
Add $2x$ on both sides of the equation,
$ \Rightarrow - 2x + 3y + 2x = 12 + 2x$
Simplify the term,
$ \Rightarrow 3y = 12 + 2x$
Divide both sides by 3,
$ \Rightarrow y = \dfrac{{12 + 2x}}{3}$...............….. (3)
Now we will put the value of y from equation (3) in equation (1),
$ \Rightarrow 2x + \dfrac{{12 + 2x}}{3} = 8$
Take LCM on the left side,
$ \Rightarrow \dfrac{{6x + 12 + 2x}}{3} = 8$
Cross multiply the terms and simplify the equation,
$ \Rightarrow 8x + 12 = 24$
Move constant on right side,
$ \Rightarrow 8x = 12$
Divide both sides by 8,
$ \Rightarrow x = \dfrac{{12}}{8}$
Cancel out the common factors,
$ \Rightarrow x = \dfrac{3}{2}$
Here we will substitute the value of x in equation (3) to get the value of y,
$ \Rightarrow y = \dfrac{{12 + 2 \times \dfrac{3}{2}}}{3}$
Cancel out the common factors,
$ \Rightarrow y = \dfrac{{12 + 3}}{3}$
Subtract the value in the numerator,
$ \Rightarrow y = \dfrac{{15}}{3}$
Divide numerator by denominator,
$ \Rightarrow y = 5$
Hence, the values of $x$ and $y$ are $\dfrac{3}{2}$ and $5$ respectively.
Note: Whenever we face such sorts of problems the key concept is to use various methods of variable evaluation either by elimination or by substitution method. These methods will help in getting the proper track to gauge these equations involving two variables and reach the proper solution.
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