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Solve the simultaneous equations
y = 9 - X
y = 2x2 + 4x + 6

User Gurwinder
by
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1 Answer

3 votes

Answer:


\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}


\begin{pmatrix}x=(1)/(2),\:&y=(17)/(2)\\ x=-3,\:&y=12\end{pmatrix}

Explanation:

Given the simultaneous equations


y=9-x


y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations


y=9-x


-


\underline{y=2x^2+4x+6}


y-y=9-x-\left(2x^2+4x+6\right)


\mathrm{Refine}


x\left(2x+5\right)=3


\mathrm{Solve\:}\:x\left(2x+5\right)=3


2x^2+5x=3
\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x


\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}


2x^2+5x-3=3-3


\mathrm{Solve\:with\:the\:quadratic\:formula}


\mathrm{Quadratic\:Equation\:Formula:}


\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}


x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)


\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_(1,\:2)=(-5\pm √(5^2-4\cdot \:2\left(-3\right)))/(2\cdot \:2)v\\


x=(-5+√(5^2-4\cdot \:2\left(-3\right)))/(2\cdot \:2)


=(-5+√(5^2+4\cdot \:2\cdot \:3))/(2\cdot \:2)


=(-5+√(49))/(2\cdot \:2)


=(-5+√(49))/(4)


=(-5+7)/(4)


=(2)/(4)


=(1)/(2)

Similarly,


x=(-5-√(5^2-4\cdot \:2\left(-3\right)))/(2\cdot \:2):\quad -3


\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}


x=(1)/(2),\:x=-3


\mathrm{Plug\:the\:solutions\:}x=(1)/(2),\:x=-3\mathrm{\:into\:}y=9-x


\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}(1)/(2):\quad y=(17)/(2)


\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12


\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}


\begin{pmatrix}x=(1)/(2),\:&y=(17)/(2)\\ x=-3,\:&y=12\end{pmatrix}

User Panshul
by
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