Answer:
The given system of equations has solutions below:
1) The solution is (2,3)
2) The solution is ([tex]\frac{-8}{7}, \frac{5}{7}[/tex])
3) The solution is infinitely many solutions
4) No solution
Step-by-step explanation:
Given system of equation are
[tex]-x+2y=4\hfill (1)[/tex]
[tex]-4x+y=-5\hfill (2)[/tex]
To solve equation by using elimination method
Multiply eqn (2) into 2
[tex]-8x+2y=-10\hfill (3)[/tex]
Now subtracting (1) and (3)
[tex]-x+2y=4[/tex]
[tex]-8x+2y=-10[/tex]
_________________
7x=14
x=[tex]\frac{14}{7}[/tex]
[tex]x=2[/tex]
Substitute x=2 in equation (1)
-2+2y=4
2y=4+2
[tex]y=\frac{6}{2}[/tex]
y=3
Therefore the solution is (2,3)
2) Given equation is
[tex]-2x+y=3\hfill (1)[/tex]
4y-4=x
Rewritting as below
[tex]x-4y=-4\hfill (2)[/tex]
To solve equation by using elimination method
multiply (2) into 2
[tex]2x-8y=-8\hfill (3)[/tex]
Adding (1) and (3)
-2x+y=3
2x-8y=-8
________
-7y=-5
[tex]y=\frac{5}{7}[/tex]
substitute [tex]y=\frac{5}{7}[/tex] in (1)
[tex]-2x+\frac{5}{7}=3[/tex]
[tex]-2x=3-\frac{5}{7}[/tex]
[tex]-2x=\frac{21-5}{7}[/tex]
[tex]x=-\frac{8}{7}[/tex]
Therefore the solution is ([tex]\frac{-8}{7},\frac{5}{7}[/tex])
3) Given equation is [tex]6x+2y=10\hfill (1)[/tex]
[tex]3x+y=5\hfill (2)[/tex]
equation (1) can be written as
2(3x+y)=10
[tex]3x+y=\frac{10}{2}[/tex]
3x+y=5
Therefore equations (1) and (2) are same therefore it has infinitely many solutions
4) Given equation is [tex]-x-2y=14\hfill (1)[/tex]
[tex]-2x-4y=12\hfill (2)[/tex]
multiply equation (1) into 2
[tex]-2x-4y=28\hfill (3)[/tex]
To solve equation by using elimination method
subtracting (2) and (3)
-2x-4y=28
-2x-4y=12
_______
[tex]28\neq -12[/tex]
therefore it has no solution
