
How do you solve the system of equation \[3x+2y=18\] and \[7x-2y=32\]?
Answer
537.9k+ views
Hint: We will assume the given equations as equation (1) and (2) respectively and use the elimination method to solve the question. Here we will eliminate the variable y by adding the two equations and find the value of x. Once the value of x is found we will substitute it in any one of the two equations given to get the value of y.
Complete step by step solution:
Here, we have been provided with the system of equation \[3x+2y=18\] and \[7x-2y=32\] and we are asked to solve it. That means we have to find the values of the variables x and y.
Here we will use the elimination method to solve the two equations as because we can see that the absolute value of the coefficient of y in both the equations is the same and the sign is different. So, we will eliminate the variable y in the first step towards the solution. After finding the value of the variable x we will substitute it in any of the two given equations and find the value of the eliminated variable y.
Let us assume the given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow 3x+2y=18\] - (1)
\[\Rightarrow 7x-2y=32\] - (2)
Adding equation (1) and (2) we get,
\[\begin{align}
& \Rightarrow \left( 3x+2y \right)+\left( 7x-2y \right)=32+18 \\
& \Rightarrow 10x=50 \\
\end{align}\]
Dividing both the sides with 10 we get,
\[\Rightarrow x=5\]
So we have obtained the value of x therefore substituting this value in equation (1) we get,
\[\begin{align}
& \Rightarrow 3\times 5+2y=18 \\
& \Rightarrow 15+2y=18 \\
& \Rightarrow 2y=18-15 \\
& \Rightarrow 2y=3 \\
\end{align}\]
Dividing both the sides with 2, we get,
$\Rightarrow y=\dfrac{3}{2}$
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 5,\dfrac{3}{2} \right)\].
Note: We can also apply the substitution method or the cross – multiplication method to solve the above question. But we may not use the cross – multiplication method unless and until mentioned in the question due to its lengthy formula which must be remembered carefully otherwise we may make the calculations wrong. Here, in the above question you can also eliminate the variable x and find the value of y first.
Complete step by step solution:
Here, we have been provided with the system of equation \[3x+2y=18\] and \[7x-2y=32\] and we are asked to solve it. That means we have to find the values of the variables x and y.
Here we will use the elimination method to solve the two equations as because we can see that the absolute value of the coefficient of y in both the equations is the same and the sign is different. So, we will eliminate the variable y in the first step towards the solution. After finding the value of the variable x we will substitute it in any of the two given equations and find the value of the eliminated variable y.
Let us assume the given equations as equation (1) and equation (2) respectively, so we have,
\[\Rightarrow 3x+2y=18\] - (1)
\[\Rightarrow 7x-2y=32\] - (2)
Adding equation (1) and (2) we get,
\[\begin{align}
& \Rightarrow \left( 3x+2y \right)+\left( 7x-2y \right)=32+18 \\
& \Rightarrow 10x=50 \\
\end{align}\]
Dividing both the sides with 10 we get,
\[\Rightarrow x=5\]
So we have obtained the value of x therefore substituting this value in equation (1) we get,
\[\begin{align}
& \Rightarrow 3\times 5+2y=18 \\
& \Rightarrow 15+2y=18 \\
& \Rightarrow 2y=18-15 \\
& \Rightarrow 2y=3 \\
\end{align}\]
Dividing both the sides with 2, we get,
$\Rightarrow y=\dfrac{3}{2}$
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 5,\dfrac{3}{2} \right)\].
Note: We can also apply the substitution method or the cross – multiplication method to solve the above question. But we may not use the cross – multiplication method unless and until mentioned in the question due to its lengthy formula which must be remembered carefully otherwise we may make the calculations wrong. Here, in the above question you can also eliminate the variable x and find the value of y first.
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