How do you solve the simultaneous equations #3x  2y = 9 # and #4x + 5y = 1/2#?
x=2, y=
Use elimination method. To eliminate x, multiply first equation by 4 and 2nd equation by 3 and subtract.
Now substituting y= 3/2 in the first equation we have 3x=9+3= 6.
This would give x= 2
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To solve the simultaneous equations (3x  2y = 9) and (4x + 5y = \frac{1}{2}), you can use the method of substitution or elimination.
Here's the solution using the method of elimination:

Multiply the first equation by (5) and the second equation by (2) to make the coefficients of (y) equal and facilitate elimination: [ 15x  10y = 45 ] [ 8x + 10y = 1 ]

Add the two equations to eliminate (y): [ 15x  10y + 8x + 10y = 45  1 ] [ 23x = 46 ]

Solve for (x): [ x = \frac{46}{23} ] [ x = 2 ]

Substitute (x = 2) into one of the original equations to find (y). Let's use the first equation: [ 3(2)  2y = 9 ] [ 6  2y = 9 ] [ 2y = 9 + 6 ] [ 2y = 3 ] [ y = \frac{3}{2} ] [ y = \frac{3}{2} ]
So, the solution to the simultaneous equations (3x  2y = 9) and (4x + 5y = \frac{1}{2}) is (x = 2) and (y = \frac{3}{2}).
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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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