
How do you solve the system \[x-4y=27\] and \[3x+y=-23\] by substitution method?
Answer
539.7k+ views
Hint: We can solve this question using basic linear equation concepts. First we have to find the value of x using any one equation. After finding the value of x we substitute that in another equation and then simplify it to find the value of y. using this value of y we find the value of x by substituting in the first equation. So that we can find both x and y values.
Complete step by step solution:
Given equations are
\[x-4y=27\]
\[3x+y=-23\]
First we have to find the value of x using any one equation.
We will take the first equation
\[x-4y=27\]
We have to keep x containing terms on LHS side and remaining terms on RHS side.
So we have to add 4y on both sides of the equation.
By adding we will get
\[\Rightarrow x-4y+4y=27+4y\]
By simplifying we will get
\[\Rightarrow x=27+4y\]
Now we have the x value so we have to substitute this in the second equation.
By substituting the x value in the second equation we will get
\[\Rightarrow 3\left( 27+4y \right)+y=-23\]
Now we have to simplify the equation.
Now we will remove the parenthesis in the equation.
By removing parenthesis we will get
\[\Rightarrow 81+12y+y=-23\]
Now we will further simplify the equation. We will get
\[\Rightarrow 81+13y=-23\]
Now we have to subtract 81 on both sides of the equation. We will get
\[\Rightarrow 81+13y-81=-23-81\]
We will get
\[\Rightarrow 13y=-104\]
Now we have to simplify it for finding the value of y.
Now we have to divide with 13 on both sides of the equation.
\[\Rightarrow \dfrac{13y}{13}=-\dfrac{104}{13}\]
By simplifying we will get
\[\Rightarrow y=-8\]
So the y value we got is \[-8\].
Now using this y value we have to find x value.
Substitute this y value in the first equation to get the value.
By substituting we will get
\[\Rightarrow x-4\left( -8 \right)=27\]
By simplifying we will get
\[\Rightarrow x+32=27\]
Now we have to subtract 32 on both sides of the equation. We will get
\[\Rightarrow x+32-32=27-32\]
\[\Rightarrow x=-5\]
So by solving the given equations we will get x and y as \[x=-5\] and \[y=8\]
Note: we can check the solution by back substituting the values in both the equations and check if they are satisfied or not. Don’t check using only one equation we have checked both the equations then only we can say the solutions are correct.
Complete step by step solution:
Given equations are
\[x-4y=27\]
\[3x+y=-23\]
First we have to find the value of x using any one equation.
We will take the first equation
\[x-4y=27\]
We have to keep x containing terms on LHS side and remaining terms on RHS side.
So we have to add 4y on both sides of the equation.
By adding we will get
\[\Rightarrow x-4y+4y=27+4y\]
By simplifying we will get
\[\Rightarrow x=27+4y\]
Now we have the x value so we have to substitute this in the second equation.
By substituting the x value in the second equation we will get
\[\Rightarrow 3\left( 27+4y \right)+y=-23\]
Now we have to simplify the equation.
Now we will remove the parenthesis in the equation.
By removing parenthesis we will get
\[\Rightarrow 81+12y+y=-23\]
Now we will further simplify the equation. We will get
\[\Rightarrow 81+13y=-23\]
Now we have to subtract 81 on both sides of the equation. We will get
\[\Rightarrow 81+13y-81=-23-81\]
We will get
\[\Rightarrow 13y=-104\]
Now we have to simplify it for finding the value of y.
Now we have to divide with 13 on both sides of the equation.
\[\Rightarrow \dfrac{13y}{13}=-\dfrac{104}{13}\]
By simplifying we will get
\[\Rightarrow y=-8\]
So the y value we got is \[-8\].
Now using this y value we have to find x value.
Substitute this y value in the first equation to get the value.
By substituting we will get
\[\Rightarrow x-4\left( -8 \right)=27\]
By simplifying we will get
\[\Rightarrow x+32=27\]
Now we have to subtract 32 on both sides of the equation. We will get
\[\Rightarrow x+32-32=27-32\]
\[\Rightarrow x=-5\]
So by solving the given equations we will get x and y as \[x=-5\] and \[y=8\]
Note: we can check the solution by back substituting the values in both the equations and check if they are satisfied or not. Don’t check using only one equation we have checked both the equations then only we can say the solutions are correct.
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