Nonlinear Systems of Equations Problem ExerciesNonlinear Systems of Equations Problem - YouTube
Nonlinear Systems of Equations Problem ExerciesNonlinear Systems of Equations Problem - YouTube
To begin with $\displaystyle (2x)^2$ is NOT equal to $\displaystyle 2x^2$.
If $\displaystyle y^2= x^2- 1$ y is NOT equal to $\displaystyle x- 1$
You seem to be having extreme difficulty with square roots! In general [itex]\sqrt{ax^2+ b[/tex] is NOT equal to ax+ b.
13. $\displaystyle y = \sqrt{x}$ , $\displaystyle y = 2x$
$\displaystyle 2x = \sqrt{x}$
$\displaystyle (2x)^2 = (\sqrt{x})^2$
$\displaystyle 4x^2 = x$
$\displaystyle 4x^2 - x = 0$
$\displaystyle x(4x-1) = 0$
$\displaystyle x = 0 \, , \, x = \frac{1}{4}$
19.
$\displaystyle x^2 + y^2 = 1$
$\displaystyle y = x $
substitute x for y ...
$\displaystyle x^2 + x^2 = 1$
$\displaystyle 2x^2 = 1$
$\displaystyle x^2 = \frac{1}{2}$
$\displaystyle x = \pm \sqrt{\frac{1}{2}}$
btw, regarding your attempt ... note that $\displaystyle \sqrt{-x^2+1} \ne -x + 1$ get that out of your head, NOW.
23.
$\displaystyle 2x^2 - y^2 = 1$
$\displaystyle x^2 - 2y^2 = -1 $
multiply the second equation by $\displaystyle -2$ to eliminate the x terms ...
$\displaystyle 2x^2 - y^2 = 1$
$\displaystyle -2x^2 + 4y^2 = 2$
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$\displaystyle 3y^2 = 3$
$\displaystyle y = \pm 1 $
it's called Latex ...
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