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horsena [70]
3 years ago
7

Solve the system by the method of elimination.

Mathematics
2 answers:
jekas [21]3 years ago
8 0

Answer:

no solution

Step-by-step explanation:

4x+3y = 6

8x + 6y = 5

Multiply the first equation by -2

-2(4x+3y) = 6*-2

-8x -6y = -12

Add this to the second equation

-8x-6y = -12

8x + 6y = 5

---------------------

0x + 0y = -7

0 = -7

Since this is never true there is no solution

nikdorinn [45]3 years ago
3 0

Answer:

X = 8/3, y= -14/9

Step-by-step explanation:

using elimination method:

subtract equation 1 from equation 2

8x-4x + 6y-3y = 5-6

4x+3y= -1

4x= -1-3y

divide both sides by 4

x = -1-3y÷4

substitute x = -1-3y/4 in equation 2

8(-1-3y)/4 +6y = 5

-8-24y/4+ 6y =5

-8-24y+6y/4 =5

-8-18y/4 = 5

Cross multy

-8-18y × 1 = 4×5

-8-18y = 20

collect like terms

-18y = 20+8

-18y = 28

divide both sides by-18

y = 28/-8

y = -14/9

put y = -14/9 in equation 1

4x+3(-14/9) = 6

4x-42/9 = 6

42/9 = 14/3

so, 4x=6+14/3

LCM =3

4x = 18+14/3

4x= 32/3

cross multiply

4x×3 = 32

12x = 32

divide both sides by 12

12x/12= 32/12

x = 8/3

so, x = 8/3, y = -14/9

check:

first equation:

4(8/3) + 3(-14/9)

32/3 - 14/3( 3 cancels 9 rem 3)

LCM= 3

32 - 14/3

= 18/3

= 6

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