How do you solve the system of equations \[5x-3y=1\] and \[9x-4y=6\]?
Answer
626.4k+ views
Hint: From the question given we have been asked to solve \[5x-3y=1\] and \[9x-4y=6\]. We can solve the above given equations by using the process of substitution. By using the substitution method, first we will make one variable subject and get its equation and substitute it in another equation and thus we can find the values of the variables x and y.
Complete step by step solution:
From the question given, it has been given that \[5x-3y=1\]
Let us assume this equation be \[\left( 1 \right)\].
\[9x-4y=6\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, let us solve the second equation, \[5x-3y=1\]
Shift from the left-hand side of the equation to the right-hand side of the equation and subtract \[1\] on both sides.
By shifting from left hand side of the equation to the right-hand side of the equation and subtracting \[1\] on both sides, we get
\[5x-1=3y\]
Then
\[\Rightarrow y=\dfrac{5x-1}{3}\]
Let it be equation \[\left( 3 \right)\]
Now, let us substitute equation \[\left( 3 \right)\] in equation \[\left( 2 \right)\].
By substituting, we get
\[9x-4y=6\]
\[\Rightarrow 9x-4\left( \dfrac{5x-1}{3} \right)=6\]
\[\Rightarrow 27x-20x+4=18\]
\[\Rightarrow 7x+4=18\]
Shift \[4\] from the left hand side of the equation to the right-hand side of the equation. By shifting, we get
\[\Rightarrow 7x=18-4\]
\[\Rightarrow 7x=14\]
\[\Rightarrow x=\dfrac{14}{7}\]
\[\Rightarrow x=2\]
Now, substitute \[x=2\] in the equation \[\left( 3 \right)\].
By substituting \[x=2\] in the equation \[\left( 3 \right)\], we get
\[\Rightarrow y=\dfrac{5x-1}{3}\]
\[\Rightarrow y=\dfrac{5\times 2-1}{3}\]
\[\Rightarrow y=\dfrac{10-1}{3}\]
\[\Rightarrow y=3\]
Therefore, the solution for the given equations is \[x=2\] and \[y=3\].
Note: We should be very careful while doing the calculation in this problem. Also, we should know all methods to solve the given equations. We can solve this question by elimination method also. For this question we have chosen a substitution method as it helps us to get the answer which can be understood easily. Like this, we have to choose their suitable method to solve the given equations. Calculation should be done very carefully while finding the solution for the given question.
Complete step by step solution:
From the question given, it has been given that \[5x-3y=1\]
Let us assume this equation be \[\left( 1 \right)\].
\[9x-4y=6\]
Let us assume this equation be \[\left( 2 \right)\].
First of all, let us solve the second equation, \[5x-3y=1\]
Shift from the left-hand side of the equation to the right-hand side of the equation and subtract \[1\] on both sides.
By shifting from left hand side of the equation to the right-hand side of the equation and subtracting \[1\] on both sides, we get
\[5x-1=3y\]
Then
\[\Rightarrow y=\dfrac{5x-1}{3}\]
Let it be equation \[\left( 3 \right)\]
Now, let us substitute equation \[\left( 3 \right)\] in equation \[\left( 2 \right)\].
By substituting, we get
\[9x-4y=6\]
\[\Rightarrow 9x-4\left( \dfrac{5x-1}{3} \right)=6\]
\[\Rightarrow 27x-20x+4=18\]
\[\Rightarrow 7x+4=18\]
Shift \[4\] from the left hand side of the equation to the right-hand side of the equation. By shifting, we get
\[\Rightarrow 7x=18-4\]
\[\Rightarrow 7x=14\]
\[\Rightarrow x=\dfrac{14}{7}\]
\[\Rightarrow x=2\]
Now, substitute \[x=2\] in the equation \[\left( 3 \right)\].
By substituting \[x=2\] in the equation \[\left( 3 \right)\], we get
\[\Rightarrow y=\dfrac{5x-1}{3}\]
\[\Rightarrow y=\dfrac{5\times 2-1}{3}\]
\[\Rightarrow y=\dfrac{10-1}{3}\]
\[\Rightarrow y=3\]
Therefore, the solution for the given equations is \[x=2\] and \[y=3\].
Note: We should be very careful while doing the calculation in this problem. Also, we should know all methods to solve the given equations. We can solve this question by elimination method also. For this question we have chosen a substitution method as it helps us to get the answer which can be understood easily. Like this, we have to choose their suitable method to solve the given equations. Calculation should be done very carefully while finding the solution for the given question.
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