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Simultaneous linear equation

Solve the following equations:


a. 3x + 5y = 8 ,

4x – 3y = 1


b. 6p + 4q = 20,

5p – 2q = 6

User Benj
by
2.9k points

2 Answers

21 votes
21 votes

Answer:

a. x = 1 / y = 1

b. p = 2 / q = 2

Explanation:

a.

3x + 5y = 8 ________( 1 )

4x - 3y = 1________ ( 2 )

( 1 ) × 3 ____ 9x + 15y = 24 _____ ( 3 )

( 2 ) × 5____ 20x - 15y = 5 _____ ( 4 )

( 3 ) + ( 4 )


9x + 15y + 20x - 15y = 24 + 5 \\ 9x + 20x + 15y - 15y = 29 \\ 29x = 29 \\ x = (29)/(29) \\ x = 1 \\

x = 1 substitute to ( 1 ) ,


3x + 5y = 8 \\ 3(1) + 5y = 8 \\ 3 + 5y = 8 \\ 5y = 8 - 3 \\ 5y = 5 \\ y = (5)/(5) \\ x = 1 \\

b.

6p + 4q = 20_______ ( 1 )

5p – 2q = 6 _______ ( 2 )

( 2 ) × 2 ____ 10p - 4q = 12 _______ ( 3 )

( 1 ) + ( 3 )


6p + 4q + 10p - 4q = 20 + 12 \\ 6p + 10p + 4q - 4q = 32 \\ 16p = 32 \\ p = (32)/(16) \\ p = 2 \\

p = 2 substitute to ( 1 ) ,


6p + 4q = 20 \\ 6(2) + 4q = 20 \\ 12 + 4q = 20 \\ 4q = 20 - 12 \\ 4q = 8 \\ q = (8)/(4) \\ q = 2 \\

User Gyro Gearless
by
3.1k points
23 votes
23 votes

Explanation:

a) 3x + 5y = 8

4x - 3y = 1

• using the elimination method:

3x + 5y = 8 (×4)

4x -3y = 1 (×3)

12x + 20y = 32

12x -9y = 3

subract 12x from both equation:

20y - - 9y= 32 -3

20y +9y = 29

29y= 29

y= 29/29

y= 1

- substituting y= 1 in :

4x - 3y= 1

4x - 3(1) = 1

4x -3 = 1

4x = 1 + 3

x = 4/4

x= 1

b) 6p+ 4q = 20

5p - 2q = 6

• using the elimination method:

6p + 4q = 20

5p - 2q = 6 (×2)

6p + 4q = 20

10p - 4q = 12

add 4q + -4q to eliminate q.

6p+ 10p = 20+12

16p = 32

p = 32/ 16

p = 2

- subtituting p = 2 in :

5p - 2q = 6

5(2) -2q =6

10 -2q = 6

-2q = 6 - 10

q = -4 / -2

q = 2

hope this helps you,

-s.

User Jmz
by
2.9k points