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 2 Solve the simultaneous equations \(2x + y = 5\) and \(x^{2} + y^{2} = 25\)

Solution

\(2x + y = 5\) (1)

\(x^{2} + y^{2} = 25\) (2)

From (1), transpose \(y\)

\[2x + y = 5\]

\[y = 5 - 2x\]

Substitute the value of \(y\) in (2)

\[x^{2} + {(5 - 2x)}^{2} = 25\]

\[{5x}^{2} - 20x = 0\]

\[5x(x - 4) = 0\]

\[x = 4\ \ or\ 0\]

Substitute the values of \(x\) in (1)

\[2x + y = 5\]

For \(x = 4\)

\[8 + y = 5\]

\[y = 5 - 8 = - 3\]

For \(x = 0\)

\[0 + y = 5\]

\[y = 5\]

The solution to the simultaneous equations \(2x + y = 5\) and \(x^{2} + y^{2} = 25\) are the ordered pairs: \((4; - 3)or\ (0;5)\)

When graphed the points where bot graphs intersect each other is the solution to the simultaneous equations.