How do you solve the following system using substitution?: #5x+2y=-4, -3x+y=20#

2 Answers
May 7, 2018

Answer:

#(x,y)to(-4,8)#

Explanation:

#5x+2y=-4to(1)#

#-3x+y=20to(2)#

#"rearrange equation "(2)" to express y in terms of x"#

#"add "3x" to both sides"#

#rArry=20+3xto(3)#

#color(blue)"substitute "y=20+3x" into equation "(1)#

#5x+2(20+3x)=-4larr"distribute"#

#rArr5x+40+6x=-4#

#rArr11x+40=-4#

#"subtract 40 from both sides"#

#rArr11x=-44#

#"divide both sides by 11"#

#rArrx=(-44)/11=-4#

#"substitute "x=-4" into equation "(3)#

#rArry=20-12=8#

#"the point of intersection "=(-4,8)#

May 7, 2018

Answer:

The solutions are #{x=-4 ; y=8}#

Explanation:

The equations are

#{(5x+2y=-4),(-3x+y=20):}#

#<=>#, #{(5x+2y=-4),(y=3x+20):}#

#<=>#, #{(5x+2(3x+20)=-4),(y=3x+20):}#

#<=>#, #{(11x+40=-4),(y=3x+20):}#

#<=>#, #{(11x=-44),(y=3x+20):}#

#<=>#, #{(x=-4),(y=3xx-4+20=8):}#