Respuesta :
Let x = carnations
Let y = roses
Let z = daisies
Total flowers = 200
And we have the following equations:
[tex]x+y+z=200[/tex]Regarding cost:
[tex]1x+2y+3z=500[/tex]20 fewer roses than daisies is given by:
[tex]y=z-20[/tex]We have a system of 3 equations, so we proceed to solve:
Substitute y = z - 20
[tex]\begin{gathered} x+z-20+z=200 \\ x+2\mleft(z-20\mright)+3z=500 \end{gathered}[/tex]Simplify
[tex]\begin{gathered} x+2z-20=200 \\ x+2z-20+20=200+20 \\ x+2z=220 \\ \text{and} \\ x+2z-40+3z=500 \\ x+5z-40=500 \\ x+5z-40+40=500+40 \\ x+5z=540 \end{gathered}[/tex]Subtract the two equations
[tex]\begin{gathered} x+5z=540 \\ - \\ x+2z=220 \\ -------- \\ 0+3z=320 \end{gathered}[/tex]Solve for z
[tex]\begin{gathered} \frac{3z}{3}=\frac{320}{3} \\ z=107 \end{gathered}[/tex]Then substitute z in y = z - 20
[tex]y=107-20=87[/tex]Next, solve for x in the first equation
[tex]\begin{gathered} x+87+107=200 \\ x+194=200 \\ x+194-194=200-194 \\ x=6 \end{gathered}[/tex]Answer:
carnations = 6
roses = 87
daisies = 107
