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need some help with math:)

Need Some Help With Math class=

Sagot :

  • x+3y=3-----(1)
  • 3y-2x=12---(2)

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]

  • x,y=(21,-18)
inspirationalgdandyt

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

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