Which pair ordered x’y is a solution to given system of linear equations 3x+y=1 5x+y=3
1 answer:
Answer:
(1, -2)
Step-by-step explanation:
Subtracting the first equation from the second gives ...
(5x +y) -(3x +y) = (3) -(1)
2x = 2 . . . . . collect terms
x = 1 . . . . . . . divide by 2
Then substituting into the first equation, we have ...
3·1 +y = 1
y = -2 . . . . . . subtract 3
The solution is (x, y) = (1, -2).
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Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2() = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.