Respuesta :

[tex]\bf \begin{cases} \quad3x+2y=-9\\ -10x+5y=-5 \end{cases}[/tex]

so hmm the idea behind the elimination, is to make the value atop or below, of either variable, the same but with a negative sign

so hmmm say... we'll eliminate "y"

so... let's see, we have 2y and 5y, both positive
what do we multiply 2y to get a negative -5y, that way, -5y +5y = 0 and poof it goes

well let's see, let's try "k"     [tex]\bf 2yk=-5y\implies k=-\cfrac{5y}{2y}\implies k=-\cfrac{5}{2}[/tex]

so.. if we multiply 2y by -5/2, we'll end up with -5y

well, let's multiply that first equation then by -5/2

[tex]\bf \begin{array}{rllll} 3x+2y=-9&\qquad\times -\frac{5}{2}\implies &-\frac{15}{2}x-5y=\frac{45}{2}\\\\ -10x+5y=-5&\implies &-10x+5y=-5\\ &&---------\\ &&\frac{-35}{2}x+0\quad=\frac{35}{2} \end{array}\\\\ -----------------------------\\\\ \cfrac{-35}{2}x=\cfrac{-35}{2}\implies \cfrac{-35x}{2}=\cfrac{-35}{2}\implies x=\cfrac{2}{-35}\cdot \cfrac{-35}{2} \\\\\\ \boxed{x=1}[/tex]

so.. now we know x = 1.... so let's plug that in  say hmmm 2nd equation

[tex]\bf -10(1)+5y=-5\implies -10+5y=-5\implies 5y=5\implies\boxed{ y=1}[/tex]

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