Respuesta :

hello

we have two equations given and we solve them simultaneously to know how many solutions it has

[tex]\begin{gathered} 9x-3y=-6\ldots eq1 \\ 5y=15x+10\ldots eq2 \\ \end{gathered}[/tex]

let's rearrange equation 2

[tex]\begin{gathered} 5y=15x+10 \\ 5y-15x=10 \end{gathered}[/tex]

now let's solve from equation 1

[tex]\begin{gathered} 9x-3y=-6 \\ \text{make x the subject} \\ 9x=-6+3y \\ \text{divide both sides by 9} \\ \frac{9x}{9}=\frac{-6+3y}{9} \\ x=-\frac{2}{3}+\frac{1}{y}\ldots eq3 \end{gathered}[/tex]

put equation 3 into equation 2

[tex]\begin{gathered} 5y-15x=10 \\ x=-\frac{2}{3}+\frac{1}{y} \\ 5y-15(\frac{-6+3y}{9})=10 \\ 5y+\frac{60-45y}{9}=10 \\ \text{take the LCM of both sides} \\ \frac{5y+60-45y}{9}=10 \\ 5y+60-45y=9\times10 \\ 60-40y=90 \\ 40y=60-90 \\ 40y=-30 \\ y=-\frac{30}{40} \\ y=-\frac{3}{4} \end{gathered}[/tex]

put y = - 3/4 into either equation 1 or 2

from equation 2

[tex]\begin{gathered} 5y-15x=10 \\ y=-\frac{3}{4} \\ 5(-\frac{3}{4})-15x=10 \\ -\frac{15}{4}-15x=10 \\ \frac{-15-15x}{4}=10 \\ -15-15x=10\times4 \\ -15-15x=40 \\ -15x=40+15 \\ -15x=55 \\ x=-\frac{55}{15} \\ x=-\frac{11}{3} \end{gathered}[/tex]

from the calculations above, the system of equations have two solutions.