# How do you solve the system 5x - y = 3 and -10x + 2y = -6?

See a solution process below:

These equations are actually the same equation. Therefore, the solution to this is they have all points in common.

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To solve the system of equations (5x - y = 3) and (-10x + 2y = -6), you can use the method of elimination or substitution. In this case, elimination is straightforward because the coefficients of (x) and (y) in the second equation are exactly twice those in the first, which makes it easy to eliminate one of the variables.

Let's use elimination to solve it:

Given equations:

- (5x - y = 3)
- (-10x + 2y = -6)

Notice that if we add twice the first equation to the second equation, the (y) terms will be eliminated. To do this, first, multiply the entire first equation by 2 to align it with the second equation's coefficients:

Multiplying the first equation by 2: [2(5x - y) = 2(3)] [10x - 2y = 6]

Now we have:

- (10x - 2y = 6)
- (-10x + 2y = -6)

Add these two equations together: [(10x - 2y) + (-10x + 2y) = 6 + (-6)] [0x + 0y = 0]

This step shows that the two equations are essentially the same line (the second is a multiple of the first), meaning there are infinitely many solutions because the lines coincide (they are the same line). Therefore, there is not a single solution to this system but rather an infinite number of solutions along the line defined by either equation.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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