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ladessa [460]
4 years ago
11

Solving each system by eliminating x-y+z= -1 x+y+3z= -3 2x-y+2z= 0

Mathematics
1 answer:
Pie4 years ago
4 0

There's only one system of equations here.

Let's eliminate x first,

x - y + z = -1

x = -1 + y - z

Substituting into  x + y + 3z = -3 gives

(-1 + y - z) + y + 3z = -3

2y + 2z = -2

y + z = -1

Substituting for x into 2x-y+2z = 0 gives

2(-1 + y - z) - y + 2z = 0

-2 + 2y - 2z - y + 2z = 0

y = 2

Got lucky on that one.  

z = -1 - y = -1 - 2 = -3

x = -1 + y - z = -1 + 2  - -3 = 4

Answer: x=4, y=2, z= -3

Check:

x-y+z = 4 - 2 + -3 = -1, good

x+y+3x = 4+2+3(-3) =  -3, good

2x-y+2x = 2(4)-2+2(-3) = 0, good

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Answer:

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Find the angle measures.
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What is the equation of the line that is perpendicular to the given line and passes though the point (3,0)? ......What is the an
Stolb23 [73]

Answer:  The correct option is (D) 5x-3y=15.

Step-by-step explanation:  We are given to find the equation of a line that is perpendicular to the graphed line and passes through the point (3, 0).

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the graphed line passes through the points (2, -1) and (-3, 2).

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m=\dfrac{d-b}{c-a}.

So, the slope of the graphed line is

m=\dfrac{2-(-1)}{-3-2}\\\\\Rightarrow m=-\dfrac{3}{5}.

We know that the product of the slopes of two perpendicular lines is -1. So, if m' is the slope of a line perpendicular to the graphed line, then

m\times m'=-1\\\\\Rightarrow -\dfrac{3}{5}\times m'=-1\\\\\Rightarrow m'=\dfrac{5}{3}.

Since the line with slope m' passes through the point (3, 0), so its equation will be

y-0=m'(x-3)\\\\\\\Rightarrow y=\dfrac{5}{3}(x-3)\\\\\Rightarrow 3y=5x-15\\\\\Rightarrow 5x-3y=15.

Thus, the required equation of the perpendicular line is 5x-3y=15.

Option (D) is CORRECT.

8 0
4 years ago
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