Answer:
x=1, y=-5
Step-by-step explanation:
Given equations are:
In order to solve the equation
Multiplying Eqn 1 by 5 and eqn 2 by 3 and subtracting them
So,
Eqn 1 becomes
15x+50y=-235
Eqn 2 becomes
15x-21y=120
Subtracting 2 from a
15x+50y - (15x-21y) = -235-120
15x+50y-15x + 21y = -355
71y = -355
y = -355/71
y =-5
Putting y= -5 in eqn 1
3x+10(-5) = -47
3x -50 = -47
3x = -47+50
3x = 3
x = 3/3
x = 1
Hence the solution is:
x=1, y=-5
Complete Question
A dog weighs 2 pounds less than 3 times the weight of a cat. The dog weighs twenty two more pounds than the cat.
Write and solve an equation to find the weights of the cat and the dog.
Answer:
Weight of cat = x = 12 pounds
Weight of Dog = y = 34 pounds
Step-by-step explanation:
Let's represent:
Weight of cat = x
Weight of Dog = y
A dog weighs 2 pounds less than 3 times the weight of a cat.
y = 3x - 2........ Equation 1
The dog weighs twenty two more pounds than the cat
y = x + 22....... Equation 2
The equation is given as:
y = y
3x - 2 = x + 22
Collect like terms
3x - x = 22 + 2
2x = 24
x = 24/2
x = 12 pounds
Solving for y using any of the equations:
y = 3x - 2
y = 3 × 12 - 2
y = 36 - 2
y = 34 pounds
Therefore:
Weight of cat = x = 12 pounds
Weight of Dog = y = 34 pounds
Answer:B At least 4 pounds but less than 6 pounds.
Step-by-step explanation:
Ok, so one way is like your way
so
if we test some numbers
for x=10
we get about 12.3/10≈1.2
for x=100000
we get 1522.03/100000≈0.01522
the top number keeps getting smaller reletive to bottom number
when x=1000000000
we get 8899.68/1000000000≈0.000009
we notice the bottom number is increaseing at a faster rate that the botom one
when x approaches infinity,the bottom number will be so big it will become a fraction that evaluates to basically zero becuase it evaluates to a very small number that is negligible
Time taken by Ria to paint a room = 4 hours
Time taken by Destiny to paint a room = 6 hours
If they work together, they complete 1 job. So,
The LCD of 4 and 6 is 12
Hence, both of them can paint the room in 
hours or 2.4 hours.
