Parallel lines and transversals

 

 

Subject
Topic
Level
Highschool
Grade
11

Recommended Lesson

 


 ks-5b55f5a50985f.jpeg 


 

Subject
Topic
Level
Highschool
Grade
10
Lesson Steps:

solved


Answer:

see worksheet

lesson image


Grade: 10
Level: Highschool
Subject: Math
Topic: Solver

Recommended Lesson

 


 bk-5b7b0b9d54294.jpeg


 


 

Subject
Topic
Level
Highschool
Grade
12

Answer

X eq 3

Y eq 9

Z eq -6

Steps

 

  1. X-2Y-3Z eq 3
  2. 3X plus Y plus Z eq 12
  3. 3X plus Y plus Z eq 15

2-3   3Y plus 5Z eq -(3)...(4)

Now (1) x 3  -3

(3x-3x) - (6y-2y) - (9z-4z) eq

                                (9-15)

-4y-5z eq -6

4y plus 5z eq 6........(5)

(4)-(5)

(3y-4y) plus (5z-5z) eq -3-6

-(y) eq -(9)

y eq 9 .............(6)

Put (6) in (4)

(3x 9) plus 5z eq -3

5z eq (-3-27) eq -30

Z eq (-30/5) eq -6......(7)

Put (6) & (7) in (1)

X eq 3 plus 2(9) plus 3(-6)

X eq 3plus 18 -18 eq 3

Hence

X eq 3

Y eq 9

Z eq -6 us-8265fec85f8427347763eee5402c1d56-tmb.jpg 

 

 

 

Recommended Lesson

  • Solve x (x-3)(x+2)=0
Subject
Topic
Level
Highschool
Grade
12

Answer

X eq 0 

X eq 3

X eq -(2)

Steps

The product of three quantities eq 0.

Means either onen of them is zero or all of them are zero.

 X eq 0

(X-3) eq 0

X eq. 3

Or (X plus 2) eq 0

X eq -(2)

 

 

Is this lesson helpful?

Angles formed by parallel lines and transversals | Geometry | Khan Academy

Parallel lines, transversal lines, corresponding angles Practice this lesson yourself on KhanAcademy.org right now: ...

Is this lesson helpful?

Course Hero

Share lesson