How do you solve 3r-5s=-35 and 2r-5s=-30?
The solution is:
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To solve the system of equations (3r - 5s = -35) and (2r - 5s = -30), you can use the method of substitution or elimination. Let's use the elimination method.
First, multiply the second equation by 3 to make the coefficients of (r) in both equations the same. This gives us (6r - 15s = -90).
Now, subtract the second equation from the first equation to eliminate (r). This gives us (3r - 6r - 5s + 15s = -35 + 90), which simplifies to (-3r + 10s = 55).
Next, solve this equation for (s), and then substitute the value of (s) back into one of the original equations to solve for (r).
Finally, substitute the values of (r) and (s) back into one of the original equations to double-check the solution.
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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.
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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