Question

If we are given a pair of straight lines $3x+4y=85$ and $6x+7y=20$. Find, ‘x’ and ‘y’.

Answer 1

Hint: We have been given two linear equations in two variables and we need to find a solution to these equations. We will use the substitution method to solve our problem. In this method, one variable is written in terms of another and is then used in the other linear equation. This gives us an equation in a single variable, thus giving us a solution. We shall proceed like this to get our answer.

Complete step-by-step solution:
We have been given the two linear equations as follows:
$\Rightarrow 3x+4y=85$
$\Rightarrow 6x+7y=20$
Here, let us write the variable ‘y’ in terms of variable ‘x’ using our first equation. This can be done as follows:
$\begin{align}
  & \Rightarrow 3x+4y=85 \\
 & \Rightarrow 4y=85-3x \\
 & \Rightarrow y=\dfrac{85-3x}{4} \\
\end{align}$
Let us say the above equation is equation number (1), so we have:
$\Rightarrow y=\dfrac{85-3x}{4}$ ....... (1)
Now, putting this value of ‘y’ in the other linear expression, we get the new equation as:
$\begin{align}
  & \Rightarrow 6x+7\left( \dfrac{85-3x}{4} \right)=20 \\
 & \Rightarrow 6x+\dfrac{595}{4}-\dfrac{21x}{4}=20 \\
 & \Rightarrow 24x+595-21x=80 \\
\end{align}$
$\begin{align}
  & \Rightarrow 3x=80-595 \\
 & \Rightarrow 3x=-515 \\
\end{align}$
$\therefore x=-\dfrac{515}{3}$
Thus, we get the value of ‘x’ as $-\dfrac{515}{3}$ . Now, that we have the value of ‘x’, putting it in equation number (1) and solving for ‘y’, we get:
$\begin{align}
  & \Rightarrow y=\dfrac{85-3\left( -\dfrac{515}{3} \right)}{4} \\
 & \Rightarrow y=\dfrac{85+515}{4} \\
 & \Rightarrow y=\dfrac{600}{4} \\
 & \therefore y=150 \\
\end{align}$
Thus, we get the value of ‘y’ as 150..
Hence, the solution set for our linear equations $3x+4y=85$ and $6x+7y=20$ comes out to be $\left( -\dfrac{515}{3},150 \right)$.

Note: A pair of linear equations is like a pair of lines and the point(s) of intersection of these lines is the solution set. There are three possible cases. In the first case, the lines are parallel but not identical, thus it has no solution. In the second case, if the parallel lines are identical, then we get an infinite solution. And in the last case, if lines are not parallel, we get one unique solution.