# arrow_back Solve the simultaneous equations $x+y=3$ and $2 x^{2}+4 x y-y=15$

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Solve the simultaneous equations $x+y=3$ and $2 x^{2}+4 x y-y=15$

\begin{array}{r}
x+y=3 \\
y=3-x \ldots \ldots \ldots \\
2 x^{2}+4 x y-y=15 \ldots \ldots \ldots . \\
\text { Substitute }(1) \text { into }(2): \\
2 x^{2}+4 x(3-x)-(3-x)=15 \\
2 x^{2}+12 x-4 x^{2}-3+x-15=0 \\
-2 x^{2}+13 x-18=0 \\
2 x^{2}-13 x+18=0 \\
(2 x-9)(x-2)=0 \\
y=-\frac{3}{2} \text { or } \quad y=1
\end{array}

OR

\begin{gathered}
x+y=3 \\
x=3-y \ldots \ldots(1) \\
2 x^{2}+4 x y-y=15 \ldots \ldots \ldots(2) \\
\text { Substitute (1) into }(2): \\
2(3-y)^{2}+4(3-y) y-y=15 \\
2 y^{2}-12 y+18-4 y^{2}+12 y-y-15=0 \\
-2 y^{2}-y+3=0 \\
2 y^{2}+y-3=0 \\
(2 y+3)(y-1)=0 \\
y=-\frac{3}{2} \text { or } \quad y=1 \\
\end{gathered}

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