The equation would be 6x+x=63. Not sure how the models play out, but good luck
Answer:
2 (real) solutions.
Step-by-step explanation:
A quadratic always has two solutions, whether they are real or complex.
Sometimes the solution is complex, involving complex numbers (2 complex), sometimes they are real and distinct (2 real), and sometimes they are real and coincident (still two real, but they are the same).
In the case of
x^2+3x = 3, or
x² + 3x -3 = 0
we apply the quadratic formula to get
x = (-3 +/- sqrt(3^2+4(1)(3))/2
to give the two solutions
{(sqrt(21)-3)/2, -(sqrt(21)+3)/2,}
both of which are real.
x + 1
__________
3x+2 I 3x² +5x -3
3x² +2x
----------
3x -3
3x+2
--------
-5 is the remainder
Answer:
yo mama
Step-by-step explanation:
lol
4x - 2y - z = - 5 ______×2x - 3y + 2z = 3 _______×13x + y - 2z = - 5 ______×1
8x - 4y - 2z = - 10x - 3y + 2z = 33x + y - 2z = - 5
8x - 4y - 2z = - 10(+) x - 3y + 2z = 3_____________9x - 7y = - 77y = 9x + 7y = 9/7x + 1
x - 3y + 2z = 3(+) 3x + y - 2z = - 5_____________4x - 2y = - 24x - 2(9/7x + 1) = - 24x - 18/7x - 2 = - 210/7x = 0x = 0
y = 9/7(0) + 1y = 1
0 - 3(1) + 2z = 32z = 6z = 3
