Question

applications of linear systems with two variables.. i need help with this problem...see attachment please

Answer 1

Are you saying you don't know how to solve when you have two equations and two unknowns.

Answer 2

i dont know how to solve with that equation... it would give me a quadratic equation right? im confused...maybe that second equation is wrong...

Answer 3

\[h=5b-3\]\[A=\frac {bh}{2}\]These equations were given to you. The first equation puts the height h in terms of b. This is then used to replace h in the second equation to get\[A=55=\frac {b(5b-3)}{2}\]So let's expand the second equation. Let's multiply both sides by 2\[110=5b^2-3b\]Rearrange and put all terms on one side to get\[5b^2-3b-110=0\] Can you understand what I did? Do you know how to solve it from here?

Answer 4

yes thats wat i had..lol but i thought it was wrong... ok and then you solve that using the quadratic formula? to get b

Answer 5

Yes.

Answer 6

can you help me from here

Answer 7

Sure. I think you dropped the 2a term somewhere.\[x=\frac{3\pm \sqrt{2209}}{10}=\frac{3\pm 47}{10}\]Can you solve it from here?

Answer 8

why is that divided by 10?

Answer 9

oh yea the 2a

Answer 10

Because the quadratic formula is\[x=\frac {-b \pm \sqrt{b^2-4ac}}{2a}\] where a=5, b=-3 and c=-110

Answer 11

That’s the fastest you can go?? right? can i substitute b now?

Answer 12

what?

Answer 13

can u use b=3+-sqr(47)/10 now to substitute?

Answer 14

No. You have the wrong expression for b. And you can also simplify the two solutions for b. One is\[b=\frac{3+47}{10}\] and the other \[b=\frac {3-47}{10}\]

Answer 15

but b should be 5...how is that going to give me 5 as the base?

Answer 16

oh yes you only use the first answer right? ok k got it

Answer 17

Right. You can't have a negative length. Did you get h?

Answer 18

yes thanks alot :)!