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If this was correctly written: Alternatively, you might have meant: 9x+2y=5 9x+2y=5

y2x+3=0 2x+3y=0

9x+2y=5 27x+6y=15

2xy=-3 -4x-6x=0

9x+2y=5 23x=15

2x=-3/y

x=15/23

9x+2y=5

2x+3/y=0 2(15/23)+3y=0

2(9x+2y)=(5) 30/23=-3y

-9(2x+3/y)=0

y=10/23, x= 15/23

18x+4y=10

-18x+-27/y=0

18x+4y=10

4y/y-10/y=27y/y

-10=-31y

y=10/31

9x+2(10/31)=5

9x=135/31

x=.048, y=10/31

Q: How do you solve this system algebracally 9x plus 2y equals 5 and y 2x plus 3 equals 0?

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