Answer:
[tex]{8, 15, 24}[/tex]
Step-by-step explanation:
In this problem we have a ratio with [tex]3[/tex] numbers
so
Let
x------> the first number
y-----> the second number
z-----> the third number
we know that
[tex]\frac{x}{y}=\frac{2}{3}[/tex] ------> equation A
[tex]\frac{x}{z}=\frac{2}{4}[/tex] ------> equation B
[tex]\frac{y}{z}=\frac{3}{4}[/tex] ------> equation C
Verify each case
case A) [tex]{8, 15, 24}[/tex]
[tex]x=8, y=15,z=24[/tex]
Substitute in the equations
[tex]\frac{x}{y}=\frac{8}{15}[/tex]
[tex]\frac{8}{15}\neq \frac{2}{3}[/tex]
therefore
The case A) not have a ratio of [tex]2:3:4[/tex]
case B) [tex]{2x,3x,4x}[/tex]
[tex]x=2x, y=3x,z=4x[/tex]
Substitute in the equations
[tex]\frac{x}{y}=\frac{2x}{3x}=\frac{2}{3}[/tex]
[tex]\frac{x}{z}=\frac{2x}{4x}=\frac{2}{4}[/tex]
[tex]\frac{y}{z}=\frac{3x}{4x}=\frac{3}{4}[/tex]
therefore
The case B) have a ratio of [tex]2:3:4[/tex]
case C) [tex]{6m^{2},9m^{2},12m^{2}}[/tex]
[tex]x=6m^{2}, y=9m^{2},z=12m^{2}[/tex]
Substitute in the equations
[tex]\frac{x}{y}=\frac{6m^{2}}{9m^{2}}=\frac{2}{3}[/tex]
[tex]\frac{x}{z}=\frac{6m^{2}}{12m^{2}}=\frac{2}{4}[/tex]
[tex]\frac{y}{z}=\frac{9m^{2}}{12m^{2}}=\frac{3}{4}[/tex]
therefore
The case C) have a ratio of [tex]2:3:4[/tex]
