Answer:
Let's solve each pair of simultaneous equations using the elimination method:
a) \(x + y = 3\) and \(x - y = 1\)
To eliminate \(y\), we add the two equations:
\[ (x + y) + (x - y) = 3 + 1 \]
\[ 2x = 4 \]
\[ x = 2 \]
Now, substitute \(x = 2\) into one of the original equations:
\[ 2 + y = 3 \]
\[ y = 3 - 2 \]
\[ y = 1 \]
Therefore, the solution is \(x = 2\) and \(y = 1\).
b) \(x + y = 5\) and \(x - y = 1\)
Adding the equations to eliminate \(y\):
\[ (x + y) + (x - y) = 5 + 1 \]
\[ 2x = 6 \]
\[ x = 3 \]
Substitute \(x = 3\) into one of the original equations:
\[ 3 + y = 5 \]
\[ y = 5 - 3 \]
\[ y = 2 \]
The solution is \(x = 3\) and \(y = 2\).
c) \(x + y = 5\) and \(x - y = 7\)
d) \(x + y = 6\) and \(x - y = 4\)
e) \(x + y = 8\) and \(x - y = 10\)
f) \(x + y = 9\) and \(x - y = 3\)
g) \(x + y = 7\) and \(2x + y = 10\)
h) \(x - y = -3\) and \(y - 2x = 1\)
i) \(2x - 3y = 1\) and \(x + 2y = 18\)
j) \(3x + 4y = 16\) and \(x + 3y = 7\)
