Solving simultaneous linear equations - SS2 Mathematics Lesson Note
Sometimes a set of simultaneous equations can have one equation be linear and the other quadratic. These equations can be solved using a substitution approach.
Example 2Ā Solve the simultaneous equationsĀ \(2x + y = 5\)Ā andĀ \(x^{2} + y^{2} = 25\)
Solution
\(2x + y = 5\)Ā (1)
\(x^{2} + y^{2} = 25\)Ā (2)
FromĀ (1), transposeĀ \(y\)
\[2x + y = 5\]
\[y = 5 - 2x\]
Substitute the value ofĀ \(y\)Ā inĀ (2)
\[x^{2} + {(5 - 2x)}^{2} = 25\]
\[{5x}^{2} - 20x = 0\]
\[5x(x - 4) = 0\]
\[x = 4\ \ or\ 0\]
Substitute the values ofĀ \(x\)Ā inĀ (1)
\[2x + y = 5\]
ForĀ \(x = 4\)
\[8 + y = 5\]
\[y = 5 - 8 = - 3\]
ForĀ \(x = 0\)
\[0 + y = 5\]
\[y = 5\]
The solution to the simultaneous equationsĀ \(2x + y = 5\)Ā andĀ \(x^{2} + y^{2} = 25\)Ā are the ordered pairs:Ā \((4; - 3)or\ (0;5)\)
When graphed the points where bot graphs intersect each other is the solution to the simultaneous equations.