What is the solution to the system of equations?
1 answer:
Answer:
- (x, y) = (8, -1)
- (x, y) = (7, 10.5)
Step-by-step explanation:
1. You can use the expression for y to substitute into the first equation:
2x +4(1/4x -3) = 12
2x +x -12 = 12 . . . . . . eliminate parentheses
3x = 24 . . . . . . . . . . . add 12; next divide by 3
x = 8 . . . . . matches choice C
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2. You can use the expression for y to substitute into the second equation:
x -2(2x -3.5) = -14
-3x +7 = -14 . . . . . . eliminate parentheses
21 = 3x . . . . . . . . . . add 3x+14; next divide by 3
7 = x . . . . . matches the third choice
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When the answer choices are sufficiently different, you only need to find one value to determine which is the correct choice. (If you want to check your work further, you can substitute the other answer value into the two equations to see if it works.)
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9514 1404 393
Answer:
possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}
actual roots: -1, (2 ±4i√2)/3
no turning points; no local extrema
end behavior is same-sign as x-value end-behavior
Step-by-step explanation:
The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.
The Rational Root Theorem tells us any rational roots will be of the form ...
±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}
= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots
Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.
The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.
The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):
f(x) = (x +1)(3x^2 -4x +12)
The complex roots of the quadratic can be found using the quadratic formula:
x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6
x = (2 ± 4i√2)/3 . . . . complex roots
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The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.
