How do you solve the system of linear equations #x + 3y = 5# and #2x  y = 5#?
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To solve the system of linear equations (x + 3y = 5) and (2x  y = 5):

Method 1: Substitution Solve one equation for one variable and substitute into the other equation: From (2x  y = 5), solve for (y): [y = 2x  5] Substitute this into (x + 3y = 5): [x + 3(2x  5) = 5] Simplify: [x + 6x  15 = 5] [7x  15 = 5] [7x = 20] [x = \frac{20}{7}]
Substitute (x) back into (y = 2x  5): [y = 2\left(\frac{20}{7}\right)  5] [y = \frac{40}{7}  \frac{35}{7}] [y = \frac{5}{7}]
So, the solution is (x = \frac{20}{7}) and (y = \frac{5}{7}).

Method 2: Elimination Multiply the second equation by 3 to eliminate (y): [3(2x  y) = 3(5)] [6x  3y = 15]
Now, we have the system: [\begin{cases} x + 3y = 5 \ 6x  3y = 15 \end{cases}]
Add the equations together to eliminate (y): [(x + 3y) + (6x  3y) = 5 + 15] [7x = 20] [x = \frac{20}{7}]
Substitute (x) into (x + 3y = 5): [\frac{20}{7} + 3y = 5] [3y = 5  \frac{20}{7}] [3y = \frac{35}{7}  \frac{20}{7}] [3y = \frac{15}{7}] [y = \frac{5}{7}]
So, the solution is (x = \frac{20}{7}) and (y = \frac{5}{7}).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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