Question

I need help with number 4

Answer 1

`\bf \begin{cases} \quad3x+2y=-9\\ -10x+5y=-5 \end{cases}`

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"
`\bf 2yk=-5y\implies k=-\cfrac{5y}{2y}\implies k=-\cfrac{5}{2}`

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


`\bf \begin{array}{rllll} 3x+2y=-9&\qquad* -(5)/(2)\implies &-(15)/(2)x-5y=(45)/(2)\\\\ -10x+5y=-5&\implies &-10x+5y=-5\\ &&---------\\ &&(-35)/(2)x+0\quad=(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}`

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


`\bf -10(1)+5y=-5\implies -10+5y=-5\implies 5y=5\implies\boxed{ y=1}`