# How do you solve #4x-3y=8# and #2x+5y=-9#?

I found:

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To solve the system of equations (4x - 3y = 8) and (2x + 5y = -9), you can use the method of substitution or elimination.

Let's use the method of elimination:

Multiply the first equation by 2 and the second equation by 4 to make the coefficients of (x) equal:

(8x - 6y = 16)

(8x + 20y = -36)

Now subtract the first equation from the second:

(8x - 8x + 20y + 6y = -36 - 16)

This simplifies to:

(26y = -52)

Divide both sides by 26:

(y = -2)

Now substitute (y = -2) into either of the original equations. Let's use the first one:

(4x - 3(-2) = 8)

(4x + 6 = 8)

(4x = 8 - 6)

(4x = 2)

(x = \frac{2}{4})

(x = \frac{1}{2})

So, the solution to the system of equations is (x = \frac{1}{2}) and (y = -2).

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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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