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What is the solution to this system of equations?
y=2x-9
3x-4y=16
A) (2-/9,4/3)
B) (3/4,-9/2)
c) (4,-1)
D) (-1,4)
To solve simultaneous equations like these, you need to combine the equations together to get a new one that has only one variable.
One way to do this would be to multiply everything in the first equation by 4, then add that to the second one. Then the y's would cancel and you would be left with 3x = 8x-20, which you can just solve normally. Then you can use the value of x with one of your given equations to find the value of y.
Alternatively, as you're given multiple choices, you can just try each of those out until you find the one that works. You shouldn't rely on that though, and it's better to be able to work it out for yourself.
Incidentally, you've kind of left yourself a clue with your multiple choices. One of them is drawing attention to itself a bit.
Why did the vector cross the road?
It wanted to be normal.
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A third way (actually just a variation of mathsyperson's first method) is to use substitution. In the first equation, you have y isolated and being equal to 2x-9. Substitute 2x-9 in for y in the second equation.
3x - 4y = 16
3x - 4(2x-9) = 16
3x - 8x + 36 =16
-5x = -20
x = 4
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