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Irina-Kira [14]
2 years ago
15

3x-6y=-12 6x+6y=30 What is the value of x and y

Mathematics
1 answer:
tatuchka [14]2 years ago
3 0

Answer:

(2,3) or x=2 and y=3

Step-by-step explanation:

find x...

3x-6y= -12

3x = 6y-12

x = 2y-4

plug x into the other equation

6(2y-4) + 6y = 30

12y-24 +6y = 30

18y = 54

y = 3

Plug y into an equation to find x

3x-6(3) = -12

3x-18= -12

3x=6

x=2

[Answer] y=3 and x=2 or (2,3)

PLEASE RATE!! I hope this helps!!

If you have any questions comment below

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victus00 [196]

Given

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you can find

\vec{a}-\vec{b}=2\vec{i}-\vec{j}+5\vec{k}-(\vec{i}+2\vec{j}+4\vec{k})=\vec{i}-3\vec{j}+\vec{k}.

Three vectors \vec{a}, \vec{b}, \vec{a}-\vec{b} form a triangle.

1.

\cos\angle 1=\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot 2+5\cdot 4}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+2^2+4^2} }=\dfrac{20}{\sqrt{30} \cdot \sqrt{21} }.

2.

\cos\angle 2=\dfrac{\vec{a}\cdot (\vec{a}-\vec{b})}{|\vec{a}|\cdot |\vec{a}-\vec{b}|}=\dfrac{2\cdot 1+(-1)\cdot (-3)+5\cdot 1}{\sqrt{2^2+(-1)^2+5^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }.

3.

\cos\angle 3=\dfrac{\vec{b}\cdot (\vec{a}-\vec{b})}{|\vec{b}|\cdot |\vec{a}-\vec{b}|}=\dfrac{1\cdot 1+2\cdot (-3)+4\cdot 1}{\sqrt{1^2+2^2+4^2}\cdot \sqrt{1^2+(-3)^2+1^2} }=\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }.

Then

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  • \angle 2=\arccos \left(\dfrac{10}{\sqrt{30} \cdot \sqrt{11} }\right)\approx 56.60^{\circ};
  • \angle 3=\arccos \left(\dfrac{-1}{\sqrt{21} \cdot \sqrt{11} }\right)\approx 93.77^{\circ}.
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