How do you solve the following system?: #4x -2y =3 , -3x -y = -6#

2 Answers
Jun 20, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve the second equation for #y#:

#-3x - y = -6#

#-3x - y + color(red)(y) + color(blue)(6) = -6 + color(blue)(6) + color(red)(y)#

#-3x - 0 + color(blue)(6) = 0 + color(red)(y)#

#-3x + 6 = y#

#y = -3x + 6#

Step 2) Substitute #(-3x + 6)# for #y# in the first equation and solve for #x#:

#4x - 2y = 3# becomes:

#4x - 2(-3x + 6) = 3#

#4x - (2 xx -3x) - (2 xx 6) = 3#

#4x - (-6x) - 12 = 3#

#4x + 6x - 12 = 3#

#(4 + 6)x - 12 = 3#

#10x - 12 = 3#

#10x - 12 + color(red)(12) = 3 + color(red)(12)#

#10x - 0 = 15#

#10x = 15#

#(10x)/color(red)(10) = 15/color(red)(10)#

#(color(red)(cancel(color(black)(10)))x)/cancel(color(red)(10)) = 15/10#

#x = 3/2#

Step 3) Substitute #3/2# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -3x + 6# becomes:

#y = (-3 xx 3/2) + 6#

#y = (-9)/2 + 6#

#y = (-9)/2 + (2/2 xx 6)#

#y = (-9)/2 + 12/2#

#y = (-9 + 12)/2#

#y = 3/2#

The Solution Is:

#x = 3/2# and #y = 3/2#

Or

#(3/2, 3/2)#

Jun 20, 2018

Answer:

#y=3/2, x=3/2#

Explanation:

#4x-2y=3-------(1)#

#-3x-y=-6-------(2)#

#(2)xx2#

#:.-6x-2y=-12------(3)#

#(1)-(3)#

#:.10x=15#

#:.x=cancel 15^3/cancel 10^2#

#:.x=3/2#

#"substitute x"=3/2 "in" (2)#

#:.-3(3/2)-y=-6#

#:.-9/2-y=-12/2#

#:.-y=-12/2+9/2#

#:.-y=-3/2#

#:.y=3/2#

~~~~~~~~~~~~~~~~
check:-

substitute #y=3/2 and x=3/2 "in" (1)#

#:.4(3/2)-2(3/2)=3#

#:.12/2-6/2=3#

#:.cancel 6^3/cancel2^1=3#

#:.3=3#