7x + 2y = 4 y = x + 1 Solve the system of equations.
do you have choices?
(1/3, 4/3) (2/9, 11/9) no solution
i think its no solution
your right
yay n.n
RyanTLopez what did you get when using substitution?
it's (2/9). (11/9)
First find x...\[7x +2y = 4\] \[7x + 2y - 4= 0\] substitute y for x... \[7x + 2(x+2) - 4 = 0\] \[9x-2 = 0\] \[9x = 2\] \[x = 2/9\]
then plug x in for x in the y equation and solve for y
then hooray :p
im not sure i understand
well... have you covered solving system of equations yet?
oh wait got it
yes
plugging in x into the y equation you get... \[y = (2/9) + 1\] which is... \[y = 2/9 + 9/9\] and you get... \[y = 11/9\]
So you're left with (2/9, 11/9)
(x,y)
The solution to your equation would be: \[x = \frac{ 2 }{ 9 }and~y=\frac{ 11 }{ 9 }\]
\(\bf 7x + 2{\color{brown}{ y}} = 4\\ {\color{brown}{ y}} = x + 1\impliedby hint \\ \quad \\ \quad \\ 7x + 2{\color{brown}{ (x+1)}} = 4\implies 7x+2x+2=4\implies 9x=2 \\ \quad \\ {\color{blue}{ x}}=\cfrac{2}{9} \\ \quad \\ 7{\color{blue}{ x}}+2{\color{brown}{ y}}=4\implies 7\left({\color{blue}{ \frac{2}{9}}}\right)+2{\color{brown}{ y}}=4\implies y=?\)
as you can see, the equivalent "y" from the 2nd equation, get SUBSTITUTED in the 1st equaiton to get "x" and then plug that back in either equation, to get "y" thus called "SUBSTITUTION" method
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