How do you solve the system \[7x-8y=112\] and \[y=-2x+9\] by substitution?
Answer
571.5k+ views
Hint: Assume the given equations as equation (1) and (2) respectively. Now, consider equation (2) where the value of y is given in terms of x. Substitute this value of y in equation (1) and solve for the value of x. Once the value of x is found, substitute it in equation (2) to find the value of y.
Complete step-by-step solution:
Here, we have been provided with the two equations: \[7x-8y=112\] and \[y=-2x+9\] and we are asked to solve it. That means we have to find the values of the variables x and y. It is also said that we have to use the substitution method.
Now, let us assume the two given equations as equation (1) and (2) respectively, so we have,
\[\Rightarrow 7x-8y=112\] - (1)
\[\Rightarrow y=-2x+9\] - (2)
Let us apply the substitution method. Here, we will substitute the value of the variable y, which is given in equation (2), in equation (1) and solve for the value of x. Once the value of x is found we will substitute this obtained value in equation (2) to get the value of y.
So, now substituting the value of y given in terms of x from equation (2) in equation (1), we get,
\[\begin{align}
& \Rightarrow 7x-8\left( -2x+9 \right)=112 \\
& \Rightarrow 7x+16x-72=112 \\
& \Rightarrow 23x=112+72 \\
& \Rightarrow 23x=184 \\
\end{align}\]
Dividing both the sides with 23 and simplifying, we get,
\[\begin{align}
& \Rightarrow \dfrac{23x}{23}=\dfrac{184}{23} \\
& \Rightarrow x=8 \\
\end{align}\]
Now, substituting the above obtained value of x in equation (2), we get,
\[\begin{align}
& \Rightarrow y=-2\left( 8 \right)+9 \\
& \Rightarrow y=-16+9 \\
& \Rightarrow y=-7 \\
\end{align}\]
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 8,-7 \right)\].
Note: One may check the answer by substituting the obtained coordinates in the two given equations. If the points satisfy both the equations then our answer is correct otherwise not. Note that we can also apply the elimination method or the cross – multiplication method to solve the system and check if we are getting the same values or not. Note that you can also substitute the value of x in terms of y from equation (2) in (1).
Complete step-by-step solution:
Here, we have been provided with the two equations: \[7x-8y=112\] and \[y=-2x+9\] and we are asked to solve it. That means we have to find the values of the variables x and y. It is also said that we have to use the substitution method.
Now, let us assume the two given equations as equation (1) and (2) respectively, so we have,
\[\Rightarrow 7x-8y=112\] - (1)
\[\Rightarrow y=-2x+9\] - (2)
Let us apply the substitution method. Here, we will substitute the value of the variable y, which is given in equation (2), in equation (1) and solve for the value of x. Once the value of x is found we will substitute this obtained value in equation (2) to get the value of y.
So, now substituting the value of y given in terms of x from equation (2) in equation (1), we get,
\[\begin{align}
& \Rightarrow 7x-8\left( -2x+9 \right)=112 \\
& \Rightarrow 7x+16x-72=112 \\
& \Rightarrow 23x=112+72 \\
& \Rightarrow 23x=184 \\
\end{align}\]
Dividing both the sides with 23 and simplifying, we get,
\[\begin{align}
& \Rightarrow \dfrac{23x}{23}=\dfrac{184}{23} \\
& \Rightarrow x=8 \\
\end{align}\]
Now, substituting the above obtained value of x in equation (2), we get,
\[\begin{align}
& \Rightarrow y=-2\left( 8 \right)+9 \\
& \Rightarrow y=-16+9 \\
& \Rightarrow y=-7 \\
\end{align}\]
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( 8,-7 \right)\].
Note: One may check the answer by substituting the obtained coordinates in the two given equations. If the points satisfy both the equations then our answer is correct otherwise not. Note that we can also apply the elimination method or the cross – multiplication method to solve the system and check if we are getting the same values or not. Note that you can also substitute the value of x in terms of y from equation (2) in (1).
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