How do you solve the following system?: #y=2x-3 , y= .5x+2 #

Answer 1

#x=10/3# and #y=11/3#.

We have the following system: #y=2x-3# #y=0.5x+2#
Notice that both equations have the same variable isolated. Since the first equaion equals y and the second one too, we conclude that: #2x-3=0.5x+2#
Now we can procced with placing all letters in one side of the equation and all numbers to the other side. #2x-0.5x=2+3# #1.5x=5# #x=5/1.5#
Simplify this fraction by multiplying both parts by 2: #x=10/3#.
Now that we have the value for x, we can replace it in any of the two equations. The result must be the same, no matter which one we choose. For the first equation: #y=2*(10/3)-3/1# #y=20/3-9/3# #y=11/3#
And for the second equation: #y=0.5*(10/3)+2/1# #y=5/3+6/3# #y=11/3#.
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Answer 2

To solve the system (y = 2x - 3) and (y = 0.5x + 2), set the equations equal to each other to find the value of (x). Then, substitute the value of (x) into one of the equations to find the corresponding value of (y).

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Answer from HIX Tutor

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.

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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