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klasskru [66]
3 years ago
6

Solve by elimination 3x+2y=19 -9x+8y=-29

Mathematics
1 answer:
Alja [10]3 years ago
4 0

Step-by-step explanation:

(3x + 2y =19)(3)

-9x + 8y =-29

9x + 6y = 57 <em><u>NOTE</u></em><em><u>:</u></em> we cross here 9x in first

-9x + 8y = -29 we cross here - 9x in second

Since 9x and - 9x are opposite

___________

6y +8y = 57 - 29

14y = 28

Y=28/14

Y=2

Replace y in second equation of this system:

-9x + 8y = - 29

-9x +8(2)= - 29

-9x + 16 = - 29

-9x = -29 - 16

-9x = - 45

X= -45/-9

X=5

Hope this helps..

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Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation  −2y=3x+7 in the form y=-\dfrac{3}{2}x-\dfrac{7}{2}. Here the slope of the given line is  m_1=-\dfrac{3}{2}. If m_2 is the slope of perpendicular line, then

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Answer 1: \dfrac{2}{3}

Part B. The slope of the line y=−2x+3 is -2. Since -\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2, then lines from part A are not parallel to line a.

Since -2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1, both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

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Answer 3: 2x+5y=-10.

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Answer 4: y=-4x+15.

Part E. Consider vectors \vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a). These vectors are collinear, then

\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.

Answer 5: -\dfrac{a}{b}=-\dfrac{d}{c}.

5 0
3 years ago
Read 2 more answers
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