Respuesta :

FLORC
let's plug in what x =, y-3 into 2x+y=12
2 (y-3)+y=12
distribute 2 to both y and -3
2y-6+y=12
add all ys
3y-6=12
+6 both sides
3y=18
÷3 both sides

y=6

plug in y into x=y-3 to find x
x=6-3
x=3
[tex]x = y - 3 [/tex] 
                                             [tex]\text{substitution}[/tex]
[tex]2x + y = 12[/tex] 

Start with the original equation  

[tex]2x+y=12 [/tex] 

[tex]2(y-3)+y=12 [/tex] 

[tex]3y-6=12 [/tex] 

Solve for y in [tex]3y-6=12 [/tex]  

[tex]3y-6=12 [/tex] 

[tex]3y=12+6 [/tex] 

[tex]3y=18 [/tex] 

[tex]y= \dfrac{18}{3} [/tex] 

[tex]y=6[/tex] 

Substitute [tex]y=6[/tex] into [tex]x=y-3[/tex]

[tex]x=y-3[/tex]
 
[tex]x=6-3 [/tex] 

[tex]x= 3[/tex]  

Therefore, [tex]y=6[/tex] and [tex]x= 3[/tex]. 

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