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mr_godi [17]
2 years ago
14

1. What is the solution of the system?

Mathematics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

Answer:

The given system of equations has solutions below:

1) The solution is (2,3)

2) The solution is (\frac{-8}{7}, \frac{5}{7})

3) The solution is infinitely many solutions

4) No solution

Step-by-step explanation:

Given system of equation are

-x+2y=4\hfill (1)

-4x+y=-5\hfill (2)

To solve equation by using elimination method

Multiply eqn (2) into 2

-8x+2y=-10\hfill (3)

Now subtracting (1) and (3)

-x+2y=4

-8x+2y=-10

_________________

7x=14

x=\frac{14}{7}

x=2

Substitute x=2 in equation (1)

-2+2y=4

2y=4+2

y=\frac{6}{2}

y=3

Therefore the solution is (2,3)

2) Given equation is

-2x+y=3\hfill (1)

4y-4=x

Rewritting as below

x-4y=-4\hfill (2)

To solve equation by using elimination method

multiply (2) into 2

2x-8y=-8\hfill (3)

Adding (1) and (3)

-2x+y=3

2x-8y=-8

________

-7y=-5

y=\frac{5}{7}

substitute y=\frac{5}{7} in (1)

-2x+\frac{5}{7}=3

-2x=3-\frac{5}{7}

-2x=\frac{21-5}{7}

x=-\frac{8}{7}

Therefore the solution is (\frac{-8}{7},\frac{5}{7})

3) Given equation is 6x+2y=10\hfill (1)

3x+y=5\hfill (2)

equation (1) can be written as

2(3x+y)=10

3x+y=\frac{10}{2}

3x+y=5

Therefore equations (1) and (2) are same therefore it has infinitely many solutions

4) Given equation is -x-2y=14\hfill (1)

-2x-4y=12\hfill (2)

multiply equation (1) into 2

-2x-4y=28\hfill (3)

To solve equation by using elimination method

subtracting (2) and (3)

-2x-4y=28

-2x-4y=12

_______

28\neq -12

therefore it has no solution

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