Answer:
One Solution:
[tex]-3x+y=7\\21x+4y=-8[/tex]
Infinitely Many Solution:
[tex]31x-y=4\\61x-2y=8[/tex]
No Solution:
[tex]y=-4r+5\\y=-4r+1[/tex]
Step-by-step explanation:
System 1
[tex]-3x+y=7\\21x+4y=-8[/tex]
[tex]Eq(1)\times7+Eq(2)\\11y=41\\y=\frac{41}{11} \\\\From\ Eq\ 1\\-3x+\frac{41}{11} =7\\-3x=7-\frac{41}{11}\\-3x=\frac{36}{11}\\x=\frac{-12}{11}[/tex]
Hence only one solution [tex](-\frac{12}{11},\frac{41}{11})[/tex]
System 2
[tex]31x-y=4\\61x-2y=8[/tex]
Both represents the same line. Hence these have infinitely many solutions.
System 3
[tex]y=-4r+5\\y=-4r+1[/tex]
Compare both the lines with [tex]y=mx+b[/tex], where m is the slope.
[tex]m=-4[/tex] for both the lines.
Hence these two lines are parallel that means there does not exists any point of intersection
So this system will not have any solution.
Given the system of equations
From 1; y = 7 + 3x
Sustitute into the equation 2:
2x- 4(7+3x) = -8
2x - 28 - 12x = -8
-10x = -8 + 28
x = -20/10
x = -2
This shows that the system of equations has one solution.
For the system of equation:
3x - y = 4
6x - 2y = 8
Since both equations are the same, hence they will have an infinite number of solutions.
For the system of equations:
y = -4r - 5
y = -4r + 1
Equate both expressions
-4r - 5 = -4r + 1
-4r+4r = 1 + 5
0 = 6
Since the resulting equation is not true hence the system of the equation will have no solution.
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