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solve 3x-3y=12 and 3x+y=8 by using substitution method

Sagot :

Tantannapolesidn
a) 3x-3y=12
  3(8)-3(4)=12
  24-12=12
  12=12

b-a) 3x+y=8
   3(1)+5=8
   3+5=8
   8=8
b-b) 3x+y=8
   3(2)+2=8
   6+2=8
   8=8
Dominiidn
[tex]Using\ Substitution \ Method; \\ \\ Let \ 3x+y=8 \ be \ the \ first \ equation. \\ Let \ 3x-3y=12 \ be \ second \ equation.\\ \\ 3x+y=8 \\ \\ y=-3x+8 \\ \\ Now \ I'll \ substitute \ "y" \ to \ the \ second \ equation \ and \ solve \ for \ "x". \\ \\ 3x-3(-3x+8)=12 \\ \\ 3x+9x+8=12 \\ \\ 3x+9x=12-8 \\ \\ \frac{12x}{12}= \frac{4}{12} \\ \\ \boxed{x= \frac{1}{3}} \\ \\ y=3\times \frac{1}{3}+8 \\ \\ \boxed{y=7} \\ \\ Thus,the \ solution \ is \ \boxed{(x,y)=( \frac{1}{3},7)} [/tex]