Question

How do you solve the system of equations `s=4r-1` and `6r-5s=-23` by algebraic method?

Answer 1

`(r,s)to(2,7)`

Explanation:

Label the equations.

`color(red)(s)=4r-1to(1)`

`6r-5color(red)(s)=-23to(2)`

Solve using the `color(blue)"substitution method"`

`"Substitute " color(red)(s)=4r-1" into equation " (2)`

`rArr6r-5(4r-1)=-23`

distribute the bracket and simplify.

`6r-20r+5=-23`

`rArr-14r+5=-23`

subtract 5 from both sides.

`-14rcancel(+5)cancel(-5)=-23-5`

`rArr-14r=-28`

divide both sides by - 14

`(cancel(-14) r)/cancel(-14)=(-28)/(-14)`

`rArrr=2`

Substitute this value into ( 1 ) to obtain the corresponding value of s

`rArrs=(4xx2)-1=8-1=7`

`rArr"solution " r=2" and " s=7`

Answer 2

`r=2,s=7`

Explanation:

Substitute the given s-value into the second formula
`6r-5*(4r-1)=-23`
Multiply, then solve
`6r-20r+5=-23`

`6r-20r=-23-5`

`-14r=-28`

`(-14r)/-14=(-28)/-14`

`r=2`

Substitute this into the first equation.
`s=4*(2)-1=8-1=7`

Check by substituting both into the second formula
`6*(2)-5*(7)=-23`
`12-35=-23`. True