How do you solve the system of equation \[3x+2y=16\] and \[2x+3y=14\]?
Answer
621.9k+ views
Hint: Assume the given equations as equation (1) and (2) respectively and use the elimination method to solve the question. To eliminate the variable y, multiply equation (1) with 3 and equation (2) with 2 and subtract the obtained equations. Cancel the variable y and find the value of x. Once the value of x is found substitute it in equation (2) to get the value of y.
Complete step by step answer:
Here, we have been provided with the system of equation \[3x+2y=16\] and \[2x+3y=14\] and we are asked to solve it. That means we have to find the values of the variables x and y.
Now, let us use the elimination method to solve the two equations. Here, we will eliminate one of the variables and find the value of another variable. After finding the value of this variable we will substitute it in any of the two given equations and find the value of the eliminated variable.
Let us assume the given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow 3x+2y=16\] - (1)
\[\Rightarrow 2x+3y=14\] - (2)
Here, we will eliminate the variable y. So, multiplying equation (1) with 3 and equation (2) with 2, we get,
\[\Rightarrow 9x+6y=48\] - (3)
\[\Rightarrow 4x+6y=28\] - (4)
Subtracting equation (4) from equation (3), we get,
\[\begin{align}
& \Rightarrow \left( 9x+6y \right)-\left( 4x+6y \right)=48-28 \\
& \Rightarrow 5x=20 \\
\end{align}\]
Dividing both the sides with 5, we get,
\[\Rightarrow x=4\]
So, we have obtained the value of x, therefore substituting this value in equation (2), we get,
\[\begin{align}
& \Rightarrow 2\times 4+3y=14 \\
& \Rightarrow 8+3y=14 \\
& \Rightarrow 3y=14-8 \\
& \Rightarrow 3y=6 \\
\end{align}\]
Dividing both the sides with 3, we get,
$\Rightarrow y=2$
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 4,2 \right)\].
Note: You may note that here the given equations are of the form ax + by = c and bx + ay = d, where c and d are constants given. As we can see that the coefficients of x and y are interchanged in the two equations, so we have a different kind of method to solve such types of questions. What we do is, we subtract the given equations to form a third equation. In the next step we add the given equations to form a fourth equation. Now, the coefficients of the variables x and y will become 1 and we can easily solve the system to get the answer.
Complete step by step answer:
Here, we have been provided with the system of equation \[3x+2y=16\] and \[2x+3y=14\] and we are asked to solve it. That means we have to find the values of the variables x and y.
Now, let us use the elimination method to solve the two equations. Here, we will eliminate one of the variables and find the value of another variable. After finding the value of this variable we will substitute it in any of the two given equations and find the value of the eliminated variable.
Let us assume the given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow 3x+2y=16\] - (1)
\[\Rightarrow 2x+3y=14\] - (2)
Here, we will eliminate the variable y. So, multiplying equation (1) with 3 and equation (2) with 2, we get,
\[\Rightarrow 9x+6y=48\] - (3)
\[\Rightarrow 4x+6y=28\] - (4)
Subtracting equation (4) from equation (3), we get,
\[\begin{align}
& \Rightarrow \left( 9x+6y \right)-\left( 4x+6y \right)=48-28 \\
& \Rightarrow 5x=20 \\
\end{align}\]
Dividing both the sides with 5, we get,
\[\Rightarrow x=4\]
So, we have obtained the value of x, therefore substituting this value in equation (2), we get,
\[\begin{align}
& \Rightarrow 2\times 4+3y=14 \\
& \Rightarrow 8+3y=14 \\
& \Rightarrow 3y=14-8 \\
& \Rightarrow 3y=6 \\
\end{align}\]
Dividing both the sides with 3, we get,
$\Rightarrow y=2$
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 4,2 \right)\].
Note: You may note that here the given equations are of the form ax + by = c and bx + ay = d, where c and d are constants given. As we can see that the coefficients of x and y are interchanged in the two equations, so we have a different kind of method to solve such types of questions. What we do is, we subtract the given equations to form a third equation. In the next step we add the given equations to form a fourth equation. Now, the coefficients of the variables x and y will become 1 and we can easily solve the system to get the answer.
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