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Question

Solve by elimination

x^2+5x+y-10=0

x^2-x-y+3=0

The solutions are (-3.14, 1,12) and (3,12,15.86)

The solutions are (-1.12, -3.14) and (-15.86, 3.12)

The solutions are (3.14, -1.12) and (15.86, 3.12)

The solutions are (1.12, 3.14) and (-3.12,15.86)

Answer 1

Ok, so for the first equation, I would subtract y from both sides, thus resulting in:

x^2+5x-10=-y

And for the second equation, I would add y to both sides, thus resulting in:

x^2-x+3=y

Then we can get rid of the y variable in the systems of equations by combination both equations:

(x^2+5x-10) + (x^2 - x +3) is the same as -y+y.

So now you have:

2x^2-4x-7=0, and you'd have to solve this equation with the quadratic formula. I hope that helped.

Answer 2

Add the two equations to eliminate y.

2x² +4x - 7 = 0

Use the quadratic equation to solve for x

x = (-4 ±√(16 - (4)(2)(-7)))/4 = 1.12, -3.12

Which are the x values for the last choice.

Checking by substituting 1.12 and -3.12 into the 2nd equation

1.12² - 1.12 - y + 3 = 0 --> y ≈ 3.14

(-3.12)² - (-3.12) - y + 3 = 0 --> y ≈ 15.86

The solutions are (1.12, 3.14) and (-3.12,15.86)

Answer 3

(x^2+5x+y-10)-(x^2-x-y+3)=0

" -x^2+x+y-3=0

6x+2y-7=0

6x+2y=7

x=0 y=3.5

1 .5

1.1666 0

There's alot of answers.

Answer 4

Hi Brianna

Given

x^2+5x+y-10=0

x^2-x-y+3=0

The equations can be combined immediately to eliminate y

2x² + 4x - 7 = 0

Then you would use the Quadratic Formula to find x

a = 2, b = 4, and c = -7

For use in

x=(−b±√(b²−4ac))∕2a

Rewrite the Equations in terms of y = x to graph them

y = -x² - 5x + 10

y = x² -x +3

Then just plug x back in to to both equations to get values for y

I got

The solutions are (1.12, 3.14) and (-3.12,15.86) as the points common to both parabolas

You can graph it at Desmos.com as well to check