y = 2/3 * x - 1 and y = 2/3 * x - 2
x - 2y = 6 and 3x - 6y = 18
y = 5x + 7 and 3y - 15x = 18
3x + 6 = y and y = - 1/3 * x - 4
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No solutions
One Solution
Infinite solutions
For the first pair of equations, we can set them equal to each other:
2/3 * x - 1 = 2/3 * x - 2
Since both sides of the equation cancel out, we are left with 0 = -1. This is not a true statement, so there are no solutions for this pair of equations.
For the second pair of equations, we can solve the first equation for x:
x - 2y = 6
x = 2y + 6
Now substitute the value of x in the second equation:
3x - 6y = 18
3(2y + 6) - 6y = 18
6y + 18 - 6y = 18
18 = 18
Since both sides of the equation cancel out, we are left with 0 = 0. This is a true statement, so there are infinite solutions for this pair of equations.
For the third pair of equations, we can solve the first equation for y:
y = 5x + 7
Now substitute the value of y in the second equation:
3y - 15x = 18
3(5x + 7) - 15x = 18
15x + 21 - 15x = 18
21 = 18
Since both sides of the equation cancel out, we are left with 3 = 0. This is not a true statement, so there are no solutions for this pair of equations.
For the fourth pair of equations, we can set them equal to each other:
3x + 6 = y
y = -1/3 * x - 4
Substitute the value of y in the first equation:
3x + 6 = -1/3 * x - 4
Multiply both sides of the equation by 3 to get rid of the fraction:
9x + 18 = -x - 12
Combine like terms:
10x = -30
Divide both sides of the equation by 10:
x = -3
Now substitute the value of x in the second equation:
y = -1/3 * (-3) - 4
y = 1 - 4
y = -3
So the solution to this pair of equations is x = -3 and y = -3, which means there is one solution.