# How do you solve the system #5x + 3y =3# and #x+ 9y = 2#?

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To solve the system of equations:

5x + 3y = 3 x + 9y = 2

You can use either substitution or elimination method. Let's use the elimination method.

- Multiply the second equation by 5 to make the coefficients of x in both equations the same:

5(x + 9y) = 5(2) 5x + 45y = 10

- Now, subtract the first equation from the modified second equation to eliminate x:

(5x + 45y) - (5x + 3y) = 10 - 3 42y = 7

- Solve for y:

y = 7/42 y = 1/6

- Substitute the value of y into one of the original equations. Let's use the first equation:

5x + 3(1/6) = 3 5x + 1/2 = 3

- Solve for x:

5x = 3 - 1/2 5x = 5/2

- Divide by 5:

x = (5/2) / 5 x = 1/2

So, the solution to the system of equations is x = 1/2 and y = 1/6.

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You can solve the system of equations (5x + 3y = 3) and (x + 9y = 2) using either the substitution method or the elimination method. Here's how to solve it using the elimination method:

Step 1: Multiply the second equation by 5 to make the coefficients of (x) in both equations equal. (5(x + 9y) = 5(2)) This simplifies to (5x + 45y = 10).

Step 2: Subtract the first equation from the modified second equation to eliminate (x). (5x + 45y - (5x + 3y) = 10 - 3) This simplifies to (42y = 7).

Step 3: Solve for (y). (y = \frac{7}{42} = \frac{1}{6}).

Step 4: Substitute the value of (y) into either of the original equations to solve for (x). Using the first equation: (5x + 3(\frac{1}{6}) = 3) This simplifies to (5x + \frac{1}{2} = 3).

Step 5: Solve for (x). (5x = 3 - \frac{1}{2} = \frac{5}{2}) (x = \frac{1}{2}).

So, the solution to the system is (x = \frac{1}{2}) and (y = \frac{1}{6}).

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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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