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igomit [66]
3 years ago
11

System of equations ( substitution methods) can someone solve these problems x=3y-5 ; x+y=3 y=x ; 3x+3y=18 y=3x+4 ; 5x-y=0

Mathematics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

(1,2)

(3,3)

(-1/2, 11/2)

Step-by-step explanation:

A system of equations is 2 or more equations for which the same x,y solution exists. We can solve the system in three ways: elimination, substitution, or graphing. Substitution can be used by substituting the value of one variable into the equation.

<u>Pair x=3y-5 and x+y=3</u>

Because x=3y-5, we can substitute 3y-5 into x in x+y=3

x+y=3

(3y-5)+y=3

3y-5+y=3

4y-5=3

4y-5+5=3+5

4y=8

y=2

We now substitute y=2 into x=3y-5.

x=3y-5

x=3(2)-5

x=6-5

x=1

<u>Pair y=x and 3x+3y=18</u>

Because y=x, we can substitute x into y in 3x+3y=18

3x+3y=18

3x+3x=18

6x=18

x=3

We now substitute x=3 into y=x.

y=3.

<u>Pair y=-3x+4 and 5x-y=0</u>

Because y=-3x+4, we can substitute -3x+4 into y in 5x-y=0

5x-y=0

5x-(-3x+4)=0

5x+3x+4=0

8x+4=0

8x=-4

x=-1/2

We now substitute x=-1/2 into y=-3x+4

y=-3x+4

y=-3(-1/2)+4

y=3/2+4

y=11/2


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Area of a Triangle
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Answer:

3.51 (round to the nearest hundreds)

Step-by-step explanation:

PART A

slope of the line passing through AC : (6-1)/(1-2) = 5/-1 = -5

equation of the line passing through B and perpendicular to AC:

y-3 = 1/5(x-3)

y-3 = 1/5x -3/5

y = 1/5 x - 3/5 + 3

y = 1/5 x + (-3+15)/5

y = 1/5 x + 12/5

PART B

line that passes through the points A and C

y-1 = -5(x-2)

y = -5x+10 + 1

y = -5x + 11

Interception of the lines

y = 1/5x + 12/5

y = -5x + 11

-5x + 11 = 1/5x + 12/5

-25x +55 = x + 12

26x = 43

x = 43/26

y = -5(43/26) + 11

y= -215/26 + 11

y = (-215+286)/26 = 71/26

D (1.65, 2.73)

PART C

AC = \sqrt{(1-2)^2 + (6-1)^2} = \sqrt{1 + 25} = \sqrt{26} = 5,09902

BD = \sqrt{(3-1.65)^2 + (3-2.73)^2} = 1,376735

PART D

AREA = (5,09902 * 1.376735)/2 = 3,509999

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