Respuesta :
For every equation, we have to choose a value for x, and solve for y.
For part a) we have that the equation is:
[tex]6x=7y[/tex]If we choose the following value for x:
[tex]x=1[/tex]And substitute it in the equation, we find the value of y:
[tex]\begin{gathered} 6(1)=7y \\ 6=7y \\ \text{Dividing both sides by 7:} \\ \frac{6}{7}=y \end{gathered}[/tex]Answer for part a) when x=1, the value of y is y=6/7
For part b) we have the equation:
[tex]5x+3y=9[/tex]If we choose the following value for x:
[tex]x=3[/tex]and substitute it in the equation to find y:
[tex]5(3)+3y=9[/tex]To solve for y, first, we solve the multiplication between 5 and 3:
[tex]15+3y=9[/tex]Now we subtract 15 to both sides:
[tex]\begin{gathered} 3y=9-15 \\ 3y=-6 \end{gathered}[/tex]Finally, divide both sides by 3:
[tex]\begin{gathered} \frac{3y}{3}=\frac{-6}{3} \\ y=-2 \end{gathered}[/tex]Answer for part b) when x=3, the value of y is y=-2
For part c) we have the equation:
[tex]y+5-\frac{1}{3}x=7[/tex]In this case, we can choose a value for x in such a way that we eliminate the fraction. For this, we can again choose the value:
[tex]x=3[/tex]And we substitute it:
[tex]y+5-\frac{1}{3}(3)=7[/tex]1/3 by 3 is equal to 1:
[tex]y+5-1=7[/tex]Next, combine the like terms on the left side 5-1 which is 4:
[tex]y+4=7[/tex]And finally, subtract 4 to both sides:
[tex]\begin{gathered} y=7-4 \\ y=3 \end{gathered}[/tex]Answer for part c) when x=3, the value of y is y=3
