What is the solution of the linear -quadratic system of equations?

Question

anonymous · Answer

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hero · Answer

$$y = x^2 + 5x -3 \y = x+2$$

Set y = y
$$x + 2 = x^2 + 5x - 3$$

Solve for x:
$$0 = x^2 + 4x - 5$$
$$0 = (x + 5)(x - 1)$$
$$x = {-5, 1}$$

Now you should be able to find y by plugging the x values into the original system.

anonymous · Answer

To do this you need to solve them simultaniously :D.

So you notice how both have y= at the start? That means we can put them equal to each other to get $$x ^{2}+5x-3=x+2$$
From here we re arrange them to get all the parts on one side.
$$x ^{2}+5x-x-3- 2=0 --> x ^{2}+4x-5=0$$

Now you solve like any other quadratic. First factorise to get (x+5)(x-1)=0. Then cancel out either constant by making x either -5 or 1 and they are your answers.

To get the y values you put the x values back into one of the original equations.

y=(-5)+2 = -3    (-5,-3)

y=(1)+2 = 3    (1,3)