To solve this system of equations, we can use either elimination or substitution method.
Elimination Method:
1. Multiply the first equation by 2 to get rid of the y variable:
10x - 2y = 30
2. Add the second equation to the new first equation:
10x - 2y + (-10x + 2y) = 30 - 30
0 = 0
This means that the two equations are equivalent, and that there are infinitely many solutions to this system of equations.
Substitution Method:
1. Solve the first equation for y:
y = 5x - 15
2. Substitute this expression for y in the second equation:
-10x + 2(5x - 15) = -30
-10x + 10x - 30 = -30
0 = 0
Again, this means that the two equations are equivalent and that there are infinitely many solutions to this system of equations.
Therefore, the solution to this system is any point (x, y) that satisfies the equation 5x - y = 15. We can also write the solution as {(x, 5x-15) | x ∈ ℝ }
{5x-y=15
{-10x+2y=-30
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