Using elimination, the equations 3x - 4y = 12 and x + 4y = 20 can be solved to show that x = 8 and y = 3. is this true or false ?

3x - 4y = 12 (1)
x + 4y = 20 (2)

equation (2) x 3: 
3x + 12y = 60 ---(2a)

(2a) - (1):
12y - (-4y) = 60 - 12
12y +  4y = 48
16y = 48
y = 3 sub into equation (2)

x + 4(3) = 20
x + 12 = 20
x = 8

Ans: x = 8 and y = 3 (True)

Both equations have a 4y term, so we can use elimination to easily solve the system.

3x - 4y = 12
x + 4y = 20

4x + 0y = 32

Now we can solve for x.

4x = 32

x = 8

Now we plug 8 in for x in either of the original equations to find y.

3(8) - 4y = 12

24 - 4y = 12

-4y = -12

y = 3

So yes, it is true that x = 8 and y = 3.

Hope this helps :)

Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0

As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ = 9 + 72
Δ = 81

Δ>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √81)/2*2 = (3-9)/4 = -6/4 = -3/2

So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2

A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.

3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.

So 2k^2 = 9 + 3k have two rational solutions (Option B).

Hope this Helps! :)