(Solved) What is the solution of the linear-quadratic system of equations?

nurul_me0w21

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12 years ago

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

y=x^2+5x-3
y-x=2

another question if you know it is:
What are the solutions of the quadratic equation?
2x^2-16x+32=0

i dont want just answers. I actually would like to be able to understand how to do them.
It would be much appreciated!

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_coco_

wrote...

12 years ago

1.Equation 1: y=x^2+5x-3 Equation 2: y-x=2
First step is to solve equation 2 for y, so you have y=x+2
Next you can sub in your equation 2 for the y in equation 1 so you have x+2=x^2+5x-3
Now you can subtract an x and a 2 from both sides to get the left side equal to zero. That gives you 0=x^2+4x-5
Now you can factor it to 0=(x+5)(x-1)
Last, you solutions are x=-5,x=1

2. 2x^2-16x+32=0
First factor out a 2 so you have 2(x^2-8x+16)=0
The you can factor the inside to (x-4)(x-4) so you have 2(x-4)(x-4)=0
So your solution is only x=4

Hope that helps.

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julie.t

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12 years ago

y=x^2+5x-3
y-x=2

Substitute y.

x^2+5x-3-x=2

Add common factors.

x^2+4x-3=2

Subtract 2 from both sides of the equation.

x^2+4x-5=0

Quadratic Formula:

[-4±?4²-(4*1*-5)]/(2*1)=0
[-4±?16-(-20)]/2=0
[-4±?36]/2=0
[-4±6]/2=0

(-4+6)/2
2/2
x=1

(-4-6)/2
-10/2
x=-5

Answer:
x=1
or
x=-5
__________

2x^2-16x+32=0

Quadratic Formula:
(-b±?b²-4ac)/2a=0

[16±?-16²-(4*2*32)]/(2*2)=0
[16±?256-(256)]/4=0
[16±?0]/4=0
[16±0]/4=0

(16+0)/4
16/4
x=4

(16-0)/4
16/4
x=4

Answer:
x=4

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julie777

wrote...

12 years ago

second equation: y-x = 2 can be written as y= 2+x

Substituting this in the first equation,

2 + x = x^2 + 5x -3

bringing all like terms together,

2 + 3 = x^2 + 5x - x

5 = x^2 + 4x

Now, to find the value of x , equate the polynomial to zero.

x^2 + 4x -5 = 0 ---------> factors of -5 are = 5, 1 but we must satisfy the equation which has a negative 5 and a positive 4 so, the signs for our factors 5 and 1 would be (+5,-1)

x^2 + 5x - 1x -5 =0

x( x +5) - 1 (x +5) = 0

(x+5) (x-1) = 0

x = -5 or x = 1

Hope that helps!

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