patti72458

07-13-2006, 11:36 PM

3x-9y=10

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

View Full Version : elimination method

patti72458

07-13-2006, 11:36 PM

3x-9y=10

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

stapel

07-14-2006, 12:25 AM

is that correct?

There are probably many valid ways to solve this system. What you did is one of them.

Now add down, and solve the resulting equation for the value of x. Plug this value in for x in either of the original equations, and solve for the value of y.

Eliz.

There are probably many valid ways to solve this system. What you did is one of them.

Now add down, and solve the resulting equation for the value of x. Plug this value in for x in either of the original equations, and solve for the value of y.

Eliz.

Denis

07-14-2006, 04:40 AM

3x-9y=10

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

NO! 15x - 45y = 50

Since 2nd equation is 15x - 45y = 20, looks like no solution 8-)

15x-45y=20

i took 5 times the first equation

which gives me 15x+45y=50

is that correct?

NO! 15x - 45y = 50

Since 2nd equation is 15x - 45y = 20, looks like no solution 8-)

stapel

07-14-2006, 10:21 AM

NO! 15x - 45y = 50

Right you are. I'd missed the (probably accidental) sign change. Thank you for catching that.

Eliz.

Right you are. I'd missed the (probably accidental) sign change. Thank you for catching that.

Eliz.

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