Solve the simultaneous equations 3x y 10 and 2x... - SS2 Mathematics Simultaneous Linear and Quadratic Equations Question

\(3x + y = 10\) (1)

\({2x}^{2} + y^{2} = 19\) (2)

From (1), transpose \(y\),

\[3x + y = 10\]

\[y = 10 - 3x\]

In (2), substitute the value of \(y\),

\[{2x}^{2} + {(10 - 3x)}^{2} = 19\]

\[{2x}^{2} + {9x}^{2} - 60x + 100 = 19\]

\[{11x}^{2} - 60x + 100 - 19 = 0\]

\[{11x}^{2} - 60x + 81 = 0\]

\[\therefore x = 3\ or\ \frac{27}{11}\]

Substitute the values of \(x\) in (1)

\[3x + y = 10\]

For \(x = 3\)

\[9 + y = 10\]

\[y = 10 - 9 = 1\]

For \(x = \frac{27}{11} = 2.45\)

\[\frac{27}{11} + y = 10\]

\[y = 10 - \frac{27}{11}\]

\[y = 7.55\]

The solution to the simultaneous equations \(3x + y = 10\) and \({2x}^{2} + y^{2} = 19\) are the ordered pairs: \((3;1)or\ (2.45;7.55)\)