Question

t=2r+3 and sr-4t=6

Answer 1

Its called systems of engagement

Answer 2

No clue how to do this, one second let me try to teach myself.

Answer 3

Im not sure how to do this either so

Answer 4

dang

Answer 5

I cant find anything of systems of engagement

Answer 6

my bad systems of equations

Answer 7

thats what its called

Answer 8

I don't know if you are to apply t to the second equation...

\(\large\bf{sr-4(2r+3)=6}\)

weird name though....never heard of it honestly...

Answer 9

is that really sr?

Answer 10

is that the answer

Answer 11

I am asking if the equation `sr-4t=6` really is `sr`.

Answer 12

no its not this is 11th grade algebra

Answer 13

is it `r-4t=6`?

Answer 14

Well you can't have one equation with three variables if the other doesn't since this is system of equations...

I can only assume that `sr - 4t = 6` is really `r-4t=6`. So I will work from there.

We have two equations...

\(\large\bf{t=2r+3,~ r-4t=6}\)

The equation of t is already simplified to one side so we can use substitution.

\(\large\bf{r-4(2r+3)=6}\)

Simplify.

\(\large\bf{-7r-12=6}\)

Add 12 to both sides.

\(\large\bf{-7r=18}\)

Divide by -7.

\(\large\bf{r=-\frac{18}{7}}\)

You would plug this into the first equation to find t.

\(\large\bf{t=2(-\frac{18}{7})+3}\)


\(\bf{Let~me~say~that~I~am~only~assuming~due~to~not~enough~being~}\)
\(\bf{unclear~about~the~variable.....~}\)

Answer 15

oh so u figured out the problem

Answer 16

how

Answer 17

The equations you gave is `t=2r+3` and `sr-4t=6`

There seems to be a mistype (because you can't find the solution if you have one equation with more variables than the other)....

Thus I ASSUMED that `sr-4t=6`is REALLY `r-4t=6`

So I worked out `t=2r+3` and `r-4t=6`

do you understand?


Can I have a photo of the problem?

Answer 18

i understand

Answer 19

ok. If this is really how the problem looks then the explanation I gave above it correct, just work out the last part...

Answer 20

ok