How do you solve the following system: #x2y=3, 2x+3y =4 #?
The solution for the system of equations is:
Solving by elimination:
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To solve the system of equations: [ \begin{cases} x  2y = 3 \ 2x + 3y = 4 \end{cases} ]

Multiply the first equation by 2 to eliminate x: [ \begin{cases} 2x  4y = 6 \ 2x + 3y = 4 \end{cases} ]

Subtract the second equation from the first equation: [ \begin{cases} (2x  4y)  (2x + 3y) = 6  4 \ 7y = 2 \ \end{cases} ]

Solve for y: [ \begin{align*} 7y & = 2 \ y & = \frac{2}{7} \end{align*} ]

Substitute the value of y into either of the original equations to solve for x. Using the first equation: [ \begin{align*} x  2\left(\frac{2}{7}\right) & = 3 \ x + \frac{4}{7} & = 3 \ x & = 3  \frac{4}{7} \ x & = \frac{21}{7}  \frac{4}{7} \ x & = \frac{17}{7} \end{align*} ]

The solution to the system of equations is ( x = \frac{17}{7} ) and ( y = \frac{2}{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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