Question

1. What is the solution of the system?

Answer 1

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