Question

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

Answer 1

`\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}`