can someone solve this? y= -x^2+5 together with x+4y=20

Question

lgbasallote · Answer

solve by system of linear equation?

radar · Answer

x=0
y=5
is one answer
one of the equations is not linear.

radar · Answer

$$x ^{2}+y-5=0$$ The above is the first equation now multiply that by 4 getting$$4x ^{2}+4y-20=0$$ Subtract the 2nd equation (in the problem) after rearranging it to look like:  x+4y-20=0 after subtracting you have remaining:
$$3x ^{2}=0$$$$x ^{2}=0$$x=0
then y = 5

radar · Answer

@Callum is their another root?

radar · Answer

I ask because 2nd degree equations usually have two roots.

anonymous · Answer

another way is simply substitution:
$$x+4(-x^2+5) - 20 = 0$$
$$x - 4x^2 +20-20 = 0$$
$$4x^2 - x = 0$$
$$x(4x-1) = 0$$
so x = 0 or x = 1/4. So there should be a y value associated with each of these. If we let x = 0 then y = 5 by the second equation. If we let x= 1/4 then $$y=1/4(20-1/4) = 1/4(79/4) = 79/16$$. And this can be checked by putting x=1/4 into the first equation:
$$y=-1/16+5 = (-1+30)/16 = 79/16$$ So you have two pairs of solutions:
$$(0,5)$$
and $$(1/4, 79/16)$$

radar · Answer

Ah yes thanks, I hope Ilovemath sees this.

anonymous · Answer

correction: $$y=-1/16+5 = (-1+80)/16 = 79/16$$

anonymous · Answer

indeed, I think when elimination is applied to nonlinear equations there is a danger that you lose a solution...

radar · Answer

I did lol