
How do you solve the system of equations \[-6x-2y=10\] and \[2x+2y=8\]?
Answer
550.8k+ views
Hint: From the question given we have been asked to solve \[-6x-2y=10\] and \[2x+2y=8\]. We can solve the above given equations by using the process of substitution. Now let us take one equation and let us find the value of y in terms of x. Now we have to substitute this value of y in terms of x in another equation. In this way, we can find the value of x for which the system of equations can get solved. By using the value of x ,we can find the value of y.
Complete step by step solution:
From the question given, it has been given that \[-6x-2y=10\]
Let us assume this equation be \[\left( 1 \right)\].
\[2x+2y=8\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, let us solve the second equation, \[2x+2y=8\]
Take \[2\] common from the left-hand side of the equation.
By taking \[2\] common from left hand side of the equation, we get
\[\Rightarrow 2\left( x+y \right)=8\]
\[\Rightarrow \left( x+y \right)=\dfrac{8}{2}\]
\[\Rightarrow x+y=4\]
Now, shift \[x\] from left hand side to the right hand side of the equation ,we get
\[\Rightarrow y=4-x\]
Let it be equation \[\left( 3 \right)\]
Now, let us substitute equation \[\left( 3 \right)\] in equation \[\left( 1 \right)\].
By substituting, we get
\[-6x-2y=10\]
\[\Rightarrow -6x-2\left( 4-x \right)=10\]
\[\Rightarrow -6x-8+2x=10\]
\[\Rightarrow -4x-8=10\]
\[\Rightarrow -4x=10+8\]
\[\Rightarrow x=\dfrac{-18}{4}\]
\[\Rightarrow x=\dfrac{-9}{2}\]
Now, substitute \[x=\dfrac{-9}{2}\] in the equation \[\left( 3 \right)\].
By substituting \[x=\dfrac{-9}{2}\] in the equation \[\left( 3 \right)\], we get
\[\Rightarrow y=4-x\]
\[\Rightarrow y=4-\left( \dfrac{-9}{2} \right)\]
\[\Rightarrow y=4+\left( \dfrac{9}{2} \right)\]
\[\Rightarrow y=\dfrac{8+9}{2}\]
\[\Rightarrow y=\dfrac{17}{2}\]
Therefore, the solution for the given equations is \[x=\dfrac{-9}{2}\] and \[y=\dfrac{17}{2}\].
Note: This problem can be solved in another process as well. Now let us draw the graph of both systems of equations. These will intersect at one point. This one point will give the intersection point of the system of equations for which these can be satisfied.
Complete step by step solution:
From the question given, it has been given that \[-6x-2y=10\]
Let us assume this equation be \[\left( 1 \right)\].
\[2x+2y=8\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, let us solve the second equation, \[2x+2y=8\]
Take \[2\] common from the left-hand side of the equation.
By taking \[2\] common from left hand side of the equation, we get
\[\Rightarrow 2\left( x+y \right)=8\]
\[\Rightarrow \left( x+y \right)=\dfrac{8}{2}\]
\[\Rightarrow x+y=4\]
Now, shift \[x\] from left hand side to the right hand side of the equation ,we get
\[\Rightarrow y=4-x\]
Let it be equation \[\left( 3 \right)\]
Now, let us substitute equation \[\left( 3 \right)\] in equation \[\left( 1 \right)\].
By substituting, we get
\[-6x-2y=10\]
\[\Rightarrow -6x-2\left( 4-x \right)=10\]
\[\Rightarrow -6x-8+2x=10\]
\[\Rightarrow -4x-8=10\]
\[\Rightarrow -4x=10+8\]
\[\Rightarrow x=\dfrac{-18}{4}\]
\[\Rightarrow x=\dfrac{-9}{2}\]
Now, substitute \[x=\dfrac{-9}{2}\] in the equation \[\left( 3 \right)\].
By substituting \[x=\dfrac{-9}{2}\] in the equation \[\left( 3 \right)\], we get
\[\Rightarrow y=4-x\]
\[\Rightarrow y=4-\left( \dfrac{-9}{2} \right)\]
\[\Rightarrow y=4+\left( \dfrac{9}{2} \right)\]
\[\Rightarrow y=\dfrac{8+9}{2}\]
\[\Rightarrow y=\dfrac{17}{2}\]
Therefore, the solution for the given equations is \[x=\dfrac{-9}{2}\] and \[y=\dfrac{17}{2}\].
Note: This problem can be solved in another process as well. Now let us draw the graph of both systems of equations. These will intersect at one point. This one point will give the intersection point of the system of equations for which these can be satisfied.
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