a. The system that consists of y=-3x+6 and y=x^2-4x is a linear-quadratic system. How would you solve the system algebraically? Graphically?

b. Solve the system in part (a).

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Use the method of substitution to detmine when the two formulas have an intersection point.

X^2-4x=-3x+6

Solve for x by moving all terms to one side and causing the other side to be equal to zero.

X^2-7X+6=0

(X-6)(X-1)=0

.:. The system is solved when X equals 6 or 1.

`-3x+6=y` (1)

`y=x^2-4x` (2)

From (1) and (2):

`y=3(2-x)` (3)

`y=(2-x)^2-4` (4)

comparing (3) to (4):

`3(2-x)=(2-x)^2-4` (5)

set: `2-x=z` (6)

`z^2-3z-4=0`

`Delta= 9-4(-4)=25 >0`

System has two different real solution:

`z=(3+-sqrt(25))/2`

```z_1=4` `z_2=-1`

So from (6): `x_1= -2` `x_2=3`

Afterword: `y_1=12` `y_2=-3`

Graph show us:

Geomethric meaning is that the two curves: (Red line `y=x^2-4x` , blue line : `y=-3x+6` )

at the point `P(-2;12)` , and `Q(3;-3)`

Plug-in -3x + 6 on the secon equation in eplace of y.

`-3x + 6 = x^2 - 4x`

put all the terms on left side. Remember to change the sign of the term you are moving to the other side.

`-3x + 6 - x^2 + 4x = 0`

Combine like terms.

`-x^2 + x + 6 = 0`

Multiply both sides by -1 to make the leading coefficient positive.

`x^2 -x -6 = 0`

Factor the left side.

`(x - 3)(x + 2) = 0`

Equate each factor to zero, and solve for x.

`x - 3 = 0 ; x + 2 = 0`

`x = 3 ; x = -2`

Solve for y for both values of x.

We can use the first equatiop.

`y = -3(3) + 6 ; y = -3(-2) + 6`

`y = -9 + 6; y = 6 + 6`

`y = -3 ; y = 12`

Thefore the **system is {(3, -3), (-2, 12)}**.

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