Thursday

August 28, 2014

August 28, 2014

Posted by **Nick** on Thursday, March 6, 2008 at 2:53pm.

Solve the system using any algebraic method.

I'm not sure how so here is the problem I'm stuck on.

9x-5y=-30

x+2y=-3

- Math(Solve the system) -
**Rhonda**, Thursday, March 6, 2008 at 3:03pmIn the second equation, subtract the 2y to get x=-2y-3. Then plug you new value for x into the first equation:

9(-2y-3)- 5y = -30. Find the answer for y and then plug it into either equation to get the answer for x.

- Math(Solve the system) -
**Nick**, Thursday, March 6, 2008 at 3:12pmUmmm....huh?? I am more of a visual learner so maybe if it was worked out completly it would help a little more.

- Math(Solve the system) -
- Math(Solve the system) -
**DLK**, Thursday, March 6, 2008 at 3:27pmmake the second equation, x+2y = -3, into x = -3-2y and substitute -3-2y into the first equation, 9x-5y. then, you will get 9(-3-2y)-5y = -30. that should simplify to -27 - 18y - 5y = -30. That should simplify to -27 - 23y = -30. that simplifies to -23y = -3. Therefore, y = 3/23.

In the second equation, x+2y = -3, substitute 3/23 for y. (x + 2 (3/23))= -3. that simplifies to x + 6/23 = -3. therefore, x = -3 6/23.

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