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Mathematics 13 Online
OpenStudy (anonymous):

A system of equations is shown below: 6x – 5y = 5 3x + 5y = 4 The x-coordinate of the solution to this system of equations is _____.

OpenStudy (anonymous):

@horsegirl27

OpenStudy (anonymous):

@HELP-ME-PLZ

pooja195 (pooja195):

\begin{bmatrix}6x-5y=5 \\3x+5y=4\end{bmatrix}

pooja195 (pooja195):

{Multiply:3x+5y=4 by 2 \[\begin{bmatrix}6x-5y=5 \\6x+10y=8\end{bmatrix}\] \[\huge~\rm~6x+10y=8\]

OpenStudy (usukidoll):

why not use elimination method?

pooja195 (pooja195):

Thats what im using as u can see

OpenStudy (usukidoll):

6x – 5y = 5 3x + 5y = 4 if you add these equations together, we can get rid of the y's.

OpenStudy (usukidoll):

then we can solve for x

pooja195 (pooja195):

Or that works too ;P 6+3=? 5+4=?

OpenStudy (anonymous):

can yu help me more

OpenStudy (usukidoll):

so by solving for x we can place x = ? into one of the equations and solve for y xD I'm just finding the best method and seeing that -5y and 5y is already there, why not cancel them out first?

OpenStudy (usukidoll):

I want to see if you understand the concepts.. so can you please add 6x – 5y = 5 3x + 5y = 4 together and solve for x? @elmo23

OpenStudy (anonymous):

-1

OpenStudy (usukidoll):

close 9x=9 (divide both sides by 9) x = 1

OpenStudy (usukidoll):

so now we can just plug in x = 1 into one of our equations...

OpenStudy (usukidoll):

let's plug in x = 1 in here 3x + 5y = 4 so we have 3(1)+5y=4 3+5y=4

OpenStudy (usukidoll):

so we need to solve for y

OpenStudy (usukidoll):

so subtract 3 on both sides (first step) and then divide both sides by 5 (second step)

OpenStudy (anonymous):

it (1,/5)

OpenStudy (usukidoll):

3+5y=4 -3 -3 5y=1 y= 1/5 correct so x = 1 and y = 1/5 we can check that these values are correct by plugging them back into the equation

OpenStudy (anonymous):

(1, 1/5) like that

OpenStudy (usukidoll):

yeah, but the question isn't asking to put it in order paired form which is (1, 1/5) so we can just leave it as x = 1 and y = 1/5 :)

OpenStudy (anonymous):

so i need to put that as the answer ? or no

OpenStudy (usukidoll):

yes just put x = 1 and y = 1/5. I'll plug the values back in the equations to prove that we have found the right values of x and y

OpenStudy (anonymous):

thanks can u help me again

OpenStudy (usukidoll):

6x – 5y = 5 3x + 5y = 4 so for x = 1 and y = 1/5 \[6(1)-5(\frac{1}{5})=5\] \[6-(\frac{5}{5})=5\] \[6-1=5\] \[5=5\] 3x + 5y = 4 \[3(1)+5(\frac{1}{5})=4\] \[3+(\frac{5}{5})=4\] \[3+1=4\] \[4=4\]

OpenStudy (usukidoll):

yeah sure.

OpenStudy (anonymous):

Nick works two jobs to pay for college. He tutors for $15 per hour and also works as a bag boy for $8 per hour. Due to his class and study schedule, Nick is only able to work up to 20 hours per week but must earn at least $150 per week. If t represents the number of hours Nick tutors and b represents the number of hours he works as a bag boy, which system of inequalities represents this scenario? t + b greater than or equal to 20 15t + 8b = 150 t + b less than or greater to 20 15t + 8b greater than or equal to 150 t + b less than or greater to 20 15t + 8b less than or greater to 150 None of the systems shown represent this scenario.

OpenStudy (usukidoll):

uh oh. I'm not good at these word problems. @rishavraj

OpenStudy (anonymous):

oh ok im trying to get a good grade so i can get every thing good

OpenStudy (anonymous):

@pooja195

pooja195 (pooja195):

Repost it someone else will help im not good with these

OpenStudy (anonymous):

ight

OpenStudy (anonymous):

@hero

pooja195 (pooja195):

I keep getting this t+b = 20 b=bag boy hours t=tutor hours 15t+8b = 150 We know he earns %15 per hour for the tutoring job and $8 for bag boy in total he earns $150 We know this but since none match it would be none of the scienerios

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