Question

How do you find the solution of the system \[x-4y=6\] and \[3x+4y=10\]?

Answer 1

Hint: In the given question, a pair of linear equations consist of two variables which are ‘x’ and ‘y’. Hence this is the question of linear equations in two variables. A pair of linear equations in two variables can be solved using a substitution method. We need to first find the value of one variable in terms of another variable and then substitute that in one of the equations, we will get our required solution set.

Complete step by step solution:
We have given the two equations:
\[x-4y=6\]-------- (1)
\[3x+4y=10\]------- (2)
From equation (1), we obtain
\[-4y=6-x\]
\[y=-\dfrac{6-x}{4}\]-------- (3)
Substituting the value of \[y=-\dfrac{6-x}{4}\] in equation (2), we obtain
\[3x+4y=10\]
\[3x+4\left( -\dfrac{6-x}{4} \right)=10\]
Simplifying the above equation, we get
\[3x-\dfrac{24}{4}+\dfrac{4x}{4}=10\]
Simplifying the LHS of the above equation by converting into simplest form, we get
\[3x-6+x=10\]
Combining the like terms,
\[4x-6=10\]
Adding 6 to both the sides of the equation, we get
\[4x-6+6=10+6\]
Simplifying the above equation, we get
\[4x=16\]
Divide both the side of equation by 4 and converting into simplest form, we get
\[x=\dfrac{16}{4}=4\]
Therefore,
\[\Rightarrow x=4\]
Substitute the value of \[x=4\] in \[y=-\dfrac{6-x}{4}\],
\[y=-\dfrac{6-x}{4}=-\dfrac{6-4}{4}=-\dfrac{2}{4}=-\dfrac{1}{2}\]
Therefore,
\[\Rightarrow y=-\dfrac{1}{2}\]
Therefore, a pair of linear equations in two variables has a solution set of (x, y) = \[\left( 4,-\dfrac{1}{2} \right)\].

Note: A pair of linear equations in two variables have two solutions, one solution for the ‘x’ i.e. the value of ‘x’ and other solution for the ‘y’ i.e. the value of ‘y’. The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. The solution of the equation will not change if the same number is added to or subtracted from both the sides of the equation, multiplying and dividing both the sides of the equations by the same non-zero number.