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solve the system of equations.
$\begin{cases} y=2x \\ y=x^{2}-x-4 end{cases}$
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the solutions of the equations are: ( $square$ ) and ( $square$ ).
Step1: Substitute $y=2x$ into quadratic equation
$2x = x^2 - x - 4$
Step2: Rearrange to standard quadratic form
$x^2 - 3x - 4 = 0$
Step3: Factor the quadratic equation
$(x-4)(x+1) = 0$
Step4: Solve for $x$ values
$x-4=0 \implies x=4$; $x+1=0 \implies x=-1$
Step5: Find corresponding $y$ values
For $x=4$: $y=2(4)=8$; For $x=-1$: $y=2(-1)=-2$
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$(4, 8)$ and $(-1, -2)$