Answer :
[tex]-5x+y=-3 \ \ \ |\times 8 \\
3x-8y=24 \\ \\
-40x+8y=-24 \\
\underline{3x-8y=24 \ \ \ \ \ \ \ } \\
-40x+3x=24-24 \\
-37x=0 \ \ \ |\div (-37) \\
x=0 \\ \\
3x-8y=24 \\
3 \times 0 -8y=24 \\
-8y=24 \ \ \ |\div (-8) \\
y=-3 \\ \\
\boxed{(x,y)=(0,-3)}[/tex]
Where going to be solving this using the substitution method so to do that we need to first find are chunk like so add 5x to each side to your first equation to get
y=(-5x-3) now we are ready to solve now that we have are chunk, first we need to insert are chunk into the other equation like this
3x-8(-5x-3)=24 now we distribute like so 3x+40x+24=24 now we add like terms
43x+24=24 now was subtract 24 from each side like so 43x=0 now we divide by 43 0/43=0 so x=0 now we can find y by inserting are x like so
y=(-5*0-3) solve for the parentheses and that leaves us with are end answer of
[tex] \left \{ {{y=-3} \atop {x=0}} \right. [/tex] Enjoy!=)
y=(-5x-3) now we are ready to solve now that we have are chunk, first we need to insert are chunk into the other equation like this
3x-8(-5x-3)=24 now we distribute like so 3x+40x+24=24 now we add like terms
43x+24=24 now was subtract 24 from each side like so 43x=0 now we divide by 43 0/43=0 so x=0 now we can find y by inserting are x like so
y=(-5*0-3) solve for the parentheses and that leaves us with are end answer of
[tex] \left \{ {{y=-3} \atop {x=0}} \right. [/tex] Enjoy!=)
