When StartFraction 2 x squared Over y EndFraction = StartFraction w + 2 Over 4 EndFraction is solved for w, one equation is w =
1 answer:
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Answer:
- A. (x1, x2) = (1, 4)
- C. There is no solution.
- B. The solutions are of the form: (x1, x2, x3) = (4t+4, t-3, t)
Step-by-step explanation:
1. Subtract the first equation from twice the second to eliminate x2.
2(2x1 -x2) -(x1 -2x2) = 2(-2) -(-7)
3x1 = 3 . . . . . simplify
x1 = 1 . . . . . . .divide by 3
Substitute this value into the second equation.
2·1 -x2 = -2
4 -x2 = 0 . . . . . add 2
4 = x2 . . . . . . . add x2
The solution is (x1, x2) = (1, 4).
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2. Multiply the first equation by 3 and add the second equation.
3(-5x1 -3x2) +(15x1 +9x2) = 3(7) +(2)
0 = 23 . . . . . not true;
There is no solution.
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3. There are two equations in 3 unknowns, so there cannot be a unique solution. The equations are not dependent, so there will be an infinite number of solutions that can be written in terms of a single parameter (t).
Let x3 = t. Then the system of equations can be rewritten as
2x1 - 5x2 = 23 +3t
x1 - 4x2 = 16
Subtracting twice the second equation from the first, we have ...
(2x1 -5x2) -2(x1 -4x2) = (23 +3t) -2(16)
3x2 = 3t -9 . . . simplify
x2 = t -3 . . . . . divide by 3
Substituting this into the second equation above, we have ...
x1 -4(t -3) = 16
x1 = 16 +4t -12 . . . . . . add 4(t-3)
x1 = 4t +4 . . . . . . . . simplify
The solutions are (x1, x2, x3) = (4t+4, t-3, t).
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The attachment shows the first system solved by graphical means. This validates our answer.
let me lie down this prism.
Check the picture below.
so the prism is really 6 rectangles stacked up to each other at the edges, so we can simply get the area of each rectangle and sum them up.
