X + 3y = 2 -x + 2y = 3 Can somebody help me?

1 answer:

4 0

X = -1 ; y = 1

x + 3y = 2, isolate x by subtracting 3y from both sides. Now that x is alone (because x = -3y + 2) take -3y + 2 and plug it into the second equation.

-(-3y + 2) + 2y =3 ; multiply what’s in the parenthesis by -1. It now is 3y -2 + 2y =3, add the common variables ( 3y + 2y) and you now have 5y-2=3. Move the 2 to the other side by adding it to both sides, you now have 5y=5, divide both sides by 5 and y=1.

Take any equation and plug y in for 1 to get x.

For example, (in the second equation), -x+2(1)=3, multiply 2 by 1 and you’re left with -x+2=3. Subtract 2 on both sides and you have -x=1, now divide both sides by -1 and x=-1.

You can check your work by plugging in the x and/or y values into the equations

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A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

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Answer:

Approximately $58.2\; \text{nautical miles}$ (assuming that the bearing is ${\rm S$29^{\circ}$W}$ .)

Step-by-step explanation:

Let $v$ denote the speed of the ship, and let $t$ denote the duration of the trip. The magnitude of the displacement of this ship would be $v\, t$ .

Refer to the diagram attached. The direction ${\rm S$29^{\circ}$W}$ means $29^{\circ}$ west of south. Thus, start with the south direction and turn towards west (clockwise) by $29^{\circ}$ to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of $v\, t$ . The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly $29^{\circ}$ .

Since the hypotenuse is of length $v\, t$ , the dashed line segment opposite to the $\theta = 29^{\circ}$ vertex would have a length of:

$\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}$ .

Substitute in $\begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned}$ and $t = 6\; \text{hour}$ :

$\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}$ .