How do you solve the system of equations #4x = 8- 2y# and #5y + 3x = 19#?
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See a solution process below:
Solve the first equation for #y#:
#4x = 8 - 2y#
#-color(red)(8) + 4x = -color(red)(8) + 8 - 2y#
#-8 + 4x = 0 - 2y#
#-8 + 4x = -2y#
#(-8 + 4x)/color(red)(-2) = (-2y)/color(red)(-2)#
#(-8)/color(red)(-2) + (4x)/color(red)(-2) = (color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2))#
#4 - 2x = y#
#y = 4 - 2x#
Step 2) Substitute #(4 - 2x)# for #y# in the second equation and solve for #x#:
#5y + 3x = 19# becomes:
#5(4 - 2x) + 3x = 19#
#(5 * 4) - (5 * 2x) + 3x = 19#
#20 - 10x + 3x = 19#
#20 + (-10 + 3)x = 19#
#20 + (-7)x = 19#
#20 - 7x = 19#
#-color(red)(20) + 20 - 7x = -color(red)(20) + 19#
#0 - 7x = -1#
#-7x = -1#
#(-7x)/color(red)(-7) = (-1)/color(red)(-7)#
#(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 1/7#
#x = 1/7#
Step 3) Substitute #1/7# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:
#y = 4 - 2x# becomes:
#y = 4 - (2 * 1/7)#
#y = 4 - 2/7#
#y = (4 * 7/7) - 2/7#
#y = 28/7 - 2/7#
#y = 26/7#
The Solution Is: #x = 1/7# and #y = 26/7# or #(1/7, 26/7)#
#4x=8-2y#
#=>4xcolor(red)(+2y)=8-2ycolor(red)(+2y)#
#=>bb("4x + 2y= 8")#
Let this be Equation number #1.#
#=>4x+2y=8 -> 1#
#5y+3x=19#
#=>bb("3x + 5y = 19")#
Let this be Equation number #2.#
#=>bb("3x + 5y = 19 ->2")#
Now #bb("Multiply")# Equation number #2# with #2.# That is#,#
#=>3x+5y=19->**2#
#=>(3x+5y)*2=19*2#
#=>3x*2+5y*2= 38#
#=>bb("6x + 10y = 38")#
Let this be Equation number be #3.#
#=>bb("6x + 10y = 38 ->3")#
Now #bb("Multiply")# Equation number #1# with #5.# That is#,#
#=>4x+2y=8->**5#
#=>(4x+2y)*5=8*5#
#=>4x*5+2y*5=40#
#=>bb("20x + 10y = 40")#
Let this be Equation number #4.#
#=>bb("20x + 10y = 40 ->4")#
Now #bb("Subtract")# Equation number #3# from Equation number #4.#
#=>(20x+10y)-(6x+10y)=40-38#
#=>20x+10y-6x-10y=2#
#=>20x-6x+10y-10y=2#
#=>14x=2#
#=>x=2/14#
#=>x=2/(2*7)#
#x=>cancelcolor(red)2/(cancelcolor(red)2*7)#
#x=>1/7#
Now put the value of #x# in Equation number #1.#
That is,
#=> 4*1/7+2y=8#
#=>4/7+2y=8#
#=>2y=8-4/7#
In the Whole number-Fraction Subtraction, #a-b/c=(ac-b)c.# Apply this,
#=>2y=(8*7-4)/7#
#=>2y=(56-4)/7#
#=>2y=52/7#
#=>y=52/(7*2)#
#=>y=(2*26)/(2*7)#
#=>y=(cancelcolor(red)2*26)/(cancelcolor(red)2*7)#
#=>y = 26/7#
So, #x=1/7# #&# #y=26/7#
