Answer:
Part A) [tex]15\ snakes[/tex]
Part B) [tex]6\ lizards[/tex]
Part C) The zoo would need to get four more lizards to maintain the same proportion
Part D) 12 snakes and 8 lizards
Step-by-step explanation:
Part A) If there were 10 lizards ,how many snakes would be there be?
Let
x ---> the number of snakes
y ---> the number of lizards
[tex]\frac{x}{y}=\frac{3}{2}[/tex] ----> equation A
we have that
[tex]y=10\ lizards[/tex]
substitute the value of y in the equation A
[tex]\frac{x}{10}=\frac{3}{2}[/tex]
solve for x
[tex]x=10(3)/2\\x=15\ snakes[/tex]
Part B) If there were 9 snakes ,how many lizards would there be?
Let
x ---> the number of snakes
y ---> the number of lizards
[tex]\frac{x}{y}=\frac{3}{2}[/tex] ----> equation A
we have that
[tex]x=9\ snakes[/tex]
substitute the value of x in the equation A
[tex]\frac{9}{y}=\frac{3}{2}[/tex]
solve for y
[tex]y=9(2)/3\\y=6\ lizards[/tex]
Part C) If the number of snakes in the zoo is increased by 6, how many more lizards would the zoo need to get to keep the same ratio?
Let
x ---> the number of snakes
y ---> the number of lizards
[tex]\frac{x}{y}=\frac{3}{2}[/tex] ----> equation A
For [tex]x=6\ snakes[/tex]
substitute the value of x in the equation A
[tex]\frac{6}{y}=\frac{3}{2}[/tex]
solve for y
[tex]y=6(2)/3\\y=4\ lizards[/tex]
therefore
The zoo would need to get four more lizards to maintain the same proportion
Part D) If the total number of snakes and lizards at the zoo was 20, how many snakes and lizards would there be?
Let
x ---> the number of snakes
y ---> the number of lizards
[tex]\frac{x}{y}=\frac{3}{2}[/tex]
isolate the variable x
[tex]x=1.5y[/tex] ----> equation A
[tex]x+y=20[/tex] ----> equation B
solve the system by substitution
substitute equation A in equation B
[tex]1.5y+y=20[/tex]
solve for y
[tex]2.5y=20[/tex]
[tex]y=8\ lizards[/tex]
Find the value of x
[tex]x=1.5(8)=12\ snakes[/tex]
therefore
12 snakes and 8 lizards
