Answer:
By elimination method:
e.
4x - y = 0
12x - 6y = 24
We can use the method of substitution. From the first equation, we have y = 4x. We substitute this value of y into the second equation:
12x - 6(4x) = 24
12x - 24x = 24
-12x = 24
x = -2
Substituting the value of x back into the first equation:
4(-2) - y = 0
-8 - y = 0
y = -8
Therefore, the solution to the system of equations is x = -2 and y = -8.
f.
3x + 15y = 3
2x - 5y = -28
We can use the method of elimination. Multiply the first equation by 2 and the second equation by 3 to eliminate x:
6x + 30y = 6
6x - 15y = -84
Subtract the second equation from the first equation:
(6x + 30y) - (6x - 15y) = 6 - (-84)
45y = 90
y = 2
Substitute the value of y into one of the original equations:
3x + 15(2) = 3
3x + 30 = 3
3x = -27
x = -9
Therefore, the solution to the system of equations is x = -9 and y = 2.
g.
3x + 2y = 10
6x + 4y = 15
We can again use the method of elimination. Multiply the first equation by 2 to eliminate y:
6x + 4y = 20
6x + 4y = 15
Since the equations are the same, we have infinitely many solutions. This means that any values of x and y that satisfy the equation 3x + 2y = 10 will be a solution to the system. They are parallel.
h.
4x - 6y = -26
-2x + 3y = 13
Again, we can use the method of elimination. Multiply the second equation by 2 to eliminate x:
4x - 6y = -26
-4x + 6y = 26
Add the two equations:
0 = 0
Since the equations are equivalent, we also have infinitely many solutions. Any values of x and y that satisfy the equation 4x - 6y = -26 will be a solution to the system. They are the same straight line.