How do you solve the system of equations #3x - 10y = - 20# and #6x - 8y = 20#?

2 Answers
Dec 15, 2016

Answer:

#{x=10,y=5]#

Explanation:

#3x-10y=-20" "(1)#

#6x-8y=20" "(2)#

#"you can multiply the both side of equation (1) by 2"#

#color(red)2(3x-10y)=color(red)(2)(-20)#

#"rearrange "#

#6x-20y=-40" "(3)#

#"now subtract (3) from (2)"#

#6x-8y-(6x-20y)=20-(-40)#

#"rearrange the equation"#

#cancel(6x)-8y-cancel(6x)+20y=20+40#

#12y=60#

#"divide both side by 12"#

#(cancel(12)y)/cancel(12)=60/12" ; "y=60/12#

#color(green)(y=5)#

#"now use (1) or (2)"#

#3x-10*color(green)(5)=-20" "(1)#

#3x-50=-20#

#3x=-20+50#

#3x=30#

#"divide both side of equation by 3"#

#(cancel(3)x)/cancel(3)=30/3#

#x=30/3#

#x=10#

Dec 15, 2016

Answer:

Different approach

#x=10"; "x=5# as solutions to this system of equations

Explanation:

Given:

#3x-10y=-20...........................Equation(1)#
#6x-8y=20..................................Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method: use one equation to express what y is 'worth' (dependant variable). Then substitute for y in the other equation so that you can solve for #x#.

Consider Equation(1)

#10y=3x+20#

#y=3/10x+2 ...........................Equation(1_a)#

Substitute for y in #Equation(2)# using #Equation(1_a)#

#6x-8y=20" "->" "6x-8(3/10x+2)=20#

#" "6x-24/10x-16=20#

#" "36/10x=36#

#" "1/10x=1#

#" "color(purple)(x=10)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I chose equation(1) as the number would be easier to work out.

Substitute #x=10# into equation(1) giving

#3x-10y=-20" "->" "3(10)-10y=-20#

#" "-10y=-50#

#" "color(purple)(y=+5)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B