How do you solve the system of equations #s=4r-1# and #6r-5s=-23# by algebraic method?
2 Answers
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#
Explanation:
Substitute the given s-value into the second formula
Multiply, then solve
Substitute this into the first equation.
Check by substituting both into the second formula