solve the system...?

Question

3x-2y=6
5x +3y= -9



factor than solve the following equations:
4x^2 +4x+1=0





x^2-8x+16=0

Answer 1

note, given

3x - 2y = 6
5x + 3y = -9

multiply the top by 3 and the bottom by 2

9x - 6y = 18
10x + 6y = 18

add the two equations together, so

19x = 0
x = 0

plug into either of the equations

3x - 2y = 6
3(0) - 2y = 6
0 - 2y = 6
y = -3

so solution to the system is (0, -3)


given 4x^2 + 4x + 1 = 0
is a perfectly squared binomial
(2x + 1)^2 = 0
x = -1/2


given x^2 - 8x + 16 = 0
(x - 4)(x - 4) = 0 again, a perfectly squared binomial
(x - 4)^2 = 0
x = 4

Answer 2

3x - 2y = 6
5x + 3y = -9
9x - 6y = 18
10x + 6y = -18
adding,
19x = 0
3y = - 9
y = - 3

4x^2 + 4x + 1=0
(2x + 1)^2 = 0
x = - 1/2

x^2 - 8x + 16=0
(x - 4)^2 = 0
x = 4

Answer 3

solve first equation for x
3x=2y+6
x=(2/3)y + 2
substitute into 2nd equation
5[(2/3)y]+3y = -9
(10/3)y+(9/3)y = -9
(19/3)y = -9
(3/19)(19/3)y = (3/19)(-9)
y = (-27/19)
substitute into first equation
x=(2/3)(-27/19) + 2
x=(-18/19) + (38/19)
= (20/19)

4(x^2+x) = -1
4(x^2+x+1/4) = -1 + 1
4(x+1/2)^2 = 0
(x+1/2) = 0
x = -1/2

(x-4)(x-4) = 0
x = 4

Answer 4

Apply the formulas for a 2-system simultaneous equations where the first equation and the second equation is as follows.
a1x + b1y = c1
a2x + b2y = c2

To solve for x and y,
x = (c1b2 - b1c2)/(a1b2 - b1a2)
y = (a1c2 - c1a2)/(a1b2 - b1a2)

For the quadratic equations,
4x^2 + 4x + 1 = 0
(2x + 1)(2x + 1) = 0
x = -0.5

x^2 - 8x + 16 = 0
(x - 2)(x - 2) = 0
x = 2

Answer 5

ok i will give more easier sol. for your problem so u can understand it easily

4x^2 + 4x + 1 = 0
(2x +1) (2x+1)
f(x) = 1/2 so the factor is (2x+1)^2
and the value for x is 1/2



x^2-8X+16=0

(x-4)(x-4)
f(x) = 4 so factor is( x-4)^2
and the value for f(X) is 4