How do you solve the following system: #8x-3y=3, 4x + 2y = 10 #?

2 Answers
Apr 5, 2016

Answer:

#(x,y)=(9/7,17/7)#

Explanation:

Given:
[1]#color(white)("XXX")8x-3y=3#
[2]#color(white)("XXX")4x+2y=10#

If we note that the #x# term in [2] is a simple multiple of that in [1],
we have an easy way to eliminate the #x# term:
Multiply [2] by #2# and subtract the result from [1]
[1]#color(white)("XXXXXXxX")8x-3y=3#
-[2]#xx2color(white)("XX")-(underline(8x+4y=20))#
[3]#color(white)("XXXXXXXXX")-7y=-17#

Dividing both sides by #(-7)#
[4]#color(white)("XXX")y=17/7#

Substituting #(17/7)# for #y# in [2]
[5]#color(white)("XXX")4x+34/7=70/7#

[6]#color(white)("XXX")4x=36/7#

[7]#color(white)("XXX")x=9/7#

Apr 5, 2016

Answer:

#(x,y)=(9/7,17/7)#

Explanation:

Solve by elimination and substitution

#color(blue)(8x-3y=3#

#color(blue)(4x+2y=10#

We can eliminate #8x# in the first equation, by #4x# in the second equation if we multiply it with #-2# (with the whole equation) to get #-8x#

#rarr-2(4x+2y=10)#

Use distributive property:

#color(brown)(a(b+c)=ab+ac#

#(or)#

#color(brown)(a(b+c=x)=ab+ac=ax#

#rarr-8x-4y=-20#

Add both of the equations

#rarr(8x-3y=3)+(-8x-4y=-20)#

#rarr-7y=-17#

#color(green)(rArry=(-17)/-7=17/7#

Substitute the value of #y# to the second equation

#rarr4x+2(17/7=10)#

#rarr4x+34/7=10#

#rarr4x=10-34/7#

#rarr4x=36/7#

#color(green)(rArrx=36/7-:4=cancel36^9/7*1/cancel4^1=9/7#