Answer:
[tex]\left \{ {{x=5} \atop {y=7}} \right.[/tex]
Step-by-step explanation:
[tex]\left \{ {{x+y=12} \atop {3y = 4x+1}} \right.[/tex]
[tex]\left \{ {{x=12-y} \atop {3y=4x+1}} \right.[/tex]
We are going to do a replacement of x for a second equation
So instead
[tex]3y=4x+1[/tex]
in the place of x we are putting 12-y from the first equation, so
[tex]3y = 4(12-y)+1[/tex]
To leave the brackets we need to do multiplication
[tex]3y = 4*12-4*y+1[/tex]
Which is
[tex]3y = 48-4y+1[/tex]
[tex]3y = -4y+49 /+4y\\7y = 49 /:7\\y = 7[/tex]
How to determine x? From the first equation:
If the x +y = 12 then the x = 12-y so x = 12-7 = 5
We should write solution as a pair:
[tex]\left \{ {{x=5} \atop {y=7}} \right.[/tex]
