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vladimir2022 [97]
3 years ago
12

3x-2y=-39 and x+3y=31 solution using elimination

Mathematics
2 answers:
Kobotan [32]3 years ago
5 0
This is hard to write... I hope this makes sense to you...
3x-2y=-39
x+3y= 31
We want to use elimination therefor, we need to either get our x or our y to add together to get zero. 
To do this, we will multiply -3 to (x=3y=31) 
NEW EQUATION
3x-2y=-39 PLUS
-3x-9y= -93 Equals
Answer: -11y= -54 
y= 4.9
Plug in to solve for x (put new y in)
3x-2(4.9)=-39
x= -9.73 
IrinaVladis [17]3 years ago
5 0
Solving for x on 3x - 2y = -39:
First, we must add 2y to both sides.
3x - 2y + 2y = -39 + 2y
3x = 2y - 39
Next, we divide both sides by 3.
3x/3 = 2y - 39/3
Answer: 2/3y - 13

Solving for y on x+3y=31:
First, we must add -x to both of the sides.
x + 3y + -x = 31 + -x
3y = -x + 31
Next, we must divide both sides by 3. (Like last time.)
3y/3 = -x + 31/3
Answer: y = -1/3x + 31/3

I hope this helps! (And what you're looking for! :>)
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It is not a straight line distance from the park to the mall. None of the answers give you that result. And if you know what displacement is, none of the answers are really displacement either. The distance is sort of a "as the crow flies." distance. There's a stop off in the middle of town.

Method

You need to use the Pythagorean Formula twice -- once from the park to the city Center and once from the city center to the mall.

Distance from the Park to the city center.

a = 3   [distance east]

b = 4  [distance south]

c = ??

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c = sqrt(3^2 + 4^2)

c = sqrt(9 + 16)          Add

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So the distance from the park to the city center is 5 miles

Distance from City center to the mall

a = 2 miles [distance east]

b = 2 miles [distance north]

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c^2 = 2^2 + 2^2   Expand this.

c^2 = 4 + 4

c^2 = 8             Take the square root of both sides.

sqrt(c^2) = sqrt(8)

c = sqrt(8)          This is the result

c = 2.8

Answer

Total distance = 5 + 2.8 = 7.8

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