need some help with math:)
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From eq(1)
[tex]\\ \sf\longmapsto x=3-3y\dots(3)[/tex]
Putting the value in eq(2)
[tex]\\ \sf\longmapsto 3y-2(3-3y)=12[/tex]
[tex]\\ \sf\longmapsto 3y-6-6y=12[/tex]
[tex]\\ \sf\longmapsto -3y-6=12[/tex]
[tex]\\ \sf\longmapsto -3y=12+6[/tex]
[tex]\\ \sf\longmapsto -3y=18[/tex]
[tex]\\ \sf\longmapsto y=\dfrac{18}{-3}[/tex]
[tex]\\ \sf\longmapsto y=-6[/tex]
Putting value in eq(3)
[tex]\\ \sf\longmapsto x=3-3(-6)[/tex]
[tex]\\ \sf\longmapsto x=3+18[/tex]
[tex]\\ \sf\longmapsto x=21[/tex]
Answer:
x=-3 & y=2
Step-by-step explanation:
x+3y=3----(i)
3y-2x=12----(ii)
Solving eqn--(i)
x+3y=3
or, x=3-3y---(iii)
Substituting the value of x from eqn --(iii) in eqn (ii) ,we get
3y-2x=12
or, 3y-2(3-3y)=12
or, 3y-6+6y=12
or,9y=18
y=18/9=2
Substituting the value of x in eqn--(iii),
x=3-3y=3-3*2=3-6=-3