anonymous 5 years ago solve the system of equation by graphing. Then classify the system x=8,y=-6, is it a solution or infinity solution

1. anonymous

You have to plot each line and see where they intersect. Geometrically, the point of intersection is that point where both equations are satisfied. Now, x=8 means...x=8 FOR ALL y...so you have a vertical line moving down that cuts the x-axis at 8. y=-6 means FOR ALL x, y = -6...so you plot a horizontal line that cuts the y-axis at -6. Now you ask yourself, "Do these lines intersect?" The answer is 'Yes'...and they intersect at the point (8,-6).

2. anonymous

3. anonymous

ok so what would the order pairs be for the solution and would it be consistent or inconsistent and would it be dependent or independent

4. anonymous

The solution is consistent since it works for both equations. The system is independent because you cannot write one equation in terms of the other (graphically, if they were consistent, they'd lie on top of each other - they don't here).

5. anonymous

so will the solution be (8,-6)

6. anonymous

yes

7. anonymous

thank you and can I become a fan of yours and how would get in touch with you

8. anonymous

I'd love it if you were a fan. If you don't see a blue link saying, "Become a fan" next to my name above, you may have to refresh the screen.

9. anonymous

I became a fan

10. anonymous

[thumbs up]

11. anonymous

12. anonymous

If you want to refer back to me re. questions, if you leave this window and go look at another question or something, I will post something here and you should get an e-mail about it. Save that e-mail and you'll have a link back to this thread.

13. anonymous

Assuming you have e-mail notification on.

14. anonymous

Solve the equation by the elimination method 5x+5y=-11 7x-3y=13

15. anonymous

This method needs you to solve one of the equations for one of the variables, and then substitute that result into the OTHER equation to remove one of the variables. When we do this, we're find the points (x,y) that will satisfy both equations. So, take the first and solve for y:$5x+5y=-11 \rightarrow 5y=-11-5x \rightarrow y=-\frac{11}{5}-x$Now substitute this expression for y in terms of x into the OTHER equation to get

16. anonymous

$7x-3\left(- \frac{11}{5}-x \right)=13$$7x+\frac{33}{5}+3x=13$$10x=\frac{32}{5}$so$x=\frac{32}{50}$

17. anonymous

which is the same as 16/25. Now substitute this x-value into ANY of the two equations (since at this x-value, the y-value we're looking for should be the same in both equations):$5 \times \frac{16}{25}+5y=-11 \rightarrow \frac{16}{5}+5y=-11 \rightarrow 5y=-\frac{71}{5}$so $y=-\frac{71}{25}$

18. anonymous

Just let me check the numbers. The method is absolutely correct, though.

19. anonymous

The numbers are wrong. I hate writing stuff out on this thing. Let me do it on paper and scan.

20. anonymous

I have 3 choices: 1) what is the solution pair, 2)Is there an infinitely many solution 3) or there is no solution

21. anonymous

Actually, I'm right.

22. anonymous

The solution pair is $x=\frac{16}{25}, y= -\frac{71}{25}$

23. anonymous

Graph the system of inequalities, y>= -3, x>=4

24. anonymous

I got (4,4)

25. anonymous

You have to graph it. It will be a region. y>=-3 means all y greater than of equal to -3, so you start by drawing a line that includes -3 (sometimes you have to draw a dashed line, if the inequality is strict (i.e < or >, rather than <= or >=)). Then your y's lie above that, so shade it lightly (with lead pencil or something (so you can use an eraser later)). Now, x>=4 means all x that are greater than or equal to 4, so you have to draw a vertical line through x=4 and shade (lightly) everything to the right. The place where the shading from both parts meet is your solution...because this is the region where BOTH of your inequalities are satisfied. Shade that heavily as your solution (and rub out the rest, if you can).

26. anonymous

27. anonymous

Hope it all helps. I have to go now. There are some other people on here who can help too. Just post a new question in the 'Ask a question' box.

28. anonymous

ok thank u