# How do you solve the system #2x-9y=6# and #4x-3y=-7#?

The solution for the system of equations is:

Solving through elimination:

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To solve the system of equations (2x - 9y = 6) and (4x - 3y = -7), you can use the method of substitution or elimination. Let's use the substitution method.

From the first equation, solve for (x): [2x = 9y + 6] [x = \frac{9y + 6}{2}]

Now substitute this expression for (x) into the second equation: [4\left(\frac{9y + 6}{2}\right) - 3y = -7]

Simplify and solve for (y): [18y + 24 - 3y = -7] [15y = -31] [y = -\frac{31}{15}]

Now substitute the value of (y) back into either of the original equations to find (x). Let's use the first equation: [2x - 9(-\frac{31}{15}) = 6]

Solve for (x): [2x + \frac{279}{15} = 6] [2x = 6 - \frac{279}{15}] [2x = \frac{90 - 279}{15}] [2x = -\frac{189}{15}] [x = -\frac{189}{30}] [x = -\frac{63}{10}]

So, the solution to the system of equations is (x = -\frac{63}{10}) and (y = -\frac{31}{15}).

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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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