Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of \(m\) for which \(y = mx + 3\).
Step 1: Solve the pair \(2x + 3y = 11\) and \(2x - 4y = -24\)
From \(2x - 4y = -24\), express \(x\) in terms of \(y\):
\[2x = -24 + 4y \Rightarrow x = -12 + 2y.\]
Substitute this in \(2x + 3y = 11\):
\[2(-12 + 2y) + 3y = 11 \Rightarrow -24 + 4y + 3y = 11 \Rightarrow -24 + 7y = 11.\]
So
\[7y = 35 \Rightarrow y = 5.\]
Then
\[x = -12 + 2y = -12 + 10 = -2.\]
Solution of the pair: \(x = -2,\ y = 5\).
Step 2: Find \(m\) in \(y = mx + 3\)
The point \((-2, 5)\) lies on the line \(y = mx + 3\). Substitute \(x = -2\) and \(y = 5\):
\[5 = m(-2) + 3 \Rightarrow 5 = -2m + 3 \Rightarrow -2m = 2 \Rightarrow m = -1.\]
Required value: \(m = -1\).
Use substitution to solve the given pair, then plug the obtained point into \(y = mx + 3\) to compute the slope \(m\) that makes the line pass through that point.