Question

What is the solution of the following system of linear equations:
\[4x-y=-6\] and \[x-2y=-5\]

Answer 1

Hint: We are given a question in which we are given a set of equations which we have to solve for ‘x’ and ‘y’. To solve the given equations, we will use the substitution method. From the first equation, which is, \[4x-y=-6\], we will write the expression in terms of ‘x’ and we get, \[x=\dfrac{y-6}{4}\]. We will then substitute the obtained expression in terms of ‘x’ in the second equation given to us, which is, \[x-2y=-5\]. On solving the expression, we will get the value of ‘y’ which we will then put in the first equation and find the value of ‘x’ as well. Hence, we will have the values of ‘x’ and ‘y’.

Complete step-by-step answer:
According to the given question, we are given a system of equations which we have to solve and find the values of ‘x’ and ‘y’.
The equations given to us are,
\[4x-y=-6\]----(1)
\[x-2y=-5\]----(2)
We can solve the above set of equations using the substitution method.
We will first take the equation (1), which is, \[4x-y=-6\].
We will write the equation (1) in terms of ‘x’ and we get,
\[\Rightarrow 4x=y-6\]
\[\Rightarrow x=\dfrac{y-6}{4}\]----(3)
Substituting the equation (3) in equation (2), we get,
\[x-2y=-5\]
We get the new expression as,
\[\Rightarrow \left( \dfrac{y-6}{4} \right)-2y=-5\]
Now, we will take the LCM so that the expression could be further computed to find the value of ‘y’, we get,
\[\Rightarrow \dfrac{y-6}{4}-\dfrac{8y}{4}=-5\]
\[\Rightarrow \dfrac{y-6-8y}{4}=-5\]
Simplifying the expression further, we get,
\[\Rightarrow y-6-8y=-20\]
We will now separate the terms with ‘y’ and the constants, so we have,
\[\Rightarrow -6-7y=-20\]
Adding 6 to both sides of the equality, we get,
\[\Rightarrow -7y=-20+6\]
\[\Rightarrow -7y=-14\]
Cancelling out the negative signs from both the sides,
\[\Rightarrow 7y=14\]
We get the value of ‘y’ as,
\[\Rightarrow y=2\]
Now, we will substitute the value of \[y=2\] in the equation (3) and we get the value of ‘x’ as,
\[\Rightarrow x=\dfrac{2-6}{4}=\dfrac{-4}{4}=-1\]
\[\Rightarrow x=-1\]
Therefore, the value of \[x=-1\] and \[y=2\].

Note: The above solution can also be carried out using the elimination method as well. The obtained values can be checked for its true values by substituting in one of the equations and if LHS and the RHS are same, then we have the correct value else not.
That is in \[4x-y=-6\], LHS we have,
\[4x-y\]
\[\Rightarrow 4\left( -1 \right)-2\]
\[\Rightarrow -4-2=-6=RHS\]
Therefore, the values of \[x=-1\] and \[y=2\].