Systems of equations problem in algebra

Question

A dog weighs 1/8 of a cow. Their combined total is 360kg, how much does the dog weigh and how much does the cow weigh?

I got this question on my algebra test, my teacher said the answer was along the lines of $$x + 8x =9x$$ $$ \implies 360/9 = 40$$ $$360 - 40$$ $$=320$$ $$\text{The cow weighs 320 and the dog weighs 40}$$

My dad and I said that it would be $360/8 = 45$ so the dog is $45$ and the cow is $315 $

So who is correct? Do it for yourselves to make sure please.

Answer 1

Your dad's solution would be the correct one if the dog's weight were $1/8$ of the combined weight. In that case, we would have the dog's weight as $360/8=45$, and then the cow's weight would be the difference, $360-45=315$.

But the problem doesn't say that the dog's weight is $1/8$ of the combined weight of $360$; it says that the dog's weight is $1/8$ of the cow's weight, which when combined with the dog's weight is $360$. Not the same thing. Notice that $45$ is $1/7$ of $315$, not $1/8$ of it. In this case, the teacher's solution is correct.

Answer 2

Just in case this is a homework problem, I will solve a VERY similar problem.

Suppose a dog weighs $1/3$ of a statue. The combined weight of the dog and the statue is $140$. Determine the weight of the cow and of the statue.

Let $x$ denote the weight of the dog. Let $y$ denote the weight of the statue. We know that the dog weighs $1/3$ of the weight of the statue, so $x=\frac{y}{3}$. Now the combined weight of the two is $140$, so $x+y=140$. Note that $x=\frac{y}{3}$ implies $3x=y$.

Substing $y=3x$ into $x+y=140$, we see that $4x=140$. Dividing by $4$ on both sides shows that $x=\frac{140}{4}=\frac{70}{2}=35$. Now putting $x=35$ into $x+y=140$, we see that $35+y=140$, and so $y=140-35=105$.

Answer 3

OK, I will answer a general question...hopefully to help you write down the necessary equations:

Thing A weighs $\frac{\diamondsuit}{\clubsuit}$ of thing B. Thing A and thing B together weigh $\frac{\spadesuit}{\heartsuit}$. Let's assume that thing A weighs $x$ and thing B weighs $y$. The first statement tells me that:

$$ x = \frac{\diamondsuit}{\clubsuit}y $$

The second statement tells me that, added together, they weigh $\frac{\spadesuit}{\heartsuit}$:

$$ x + y = \frac{\spadesuit}{\heartsuit} $$

You can solve this through substitution: $x = \frac{\diamondsuit}{\clubsuit}y$ so plug that into the second equation to find $y$:

\begin{align} \frac{\diamondsuit}{\clubsuit}y + y = \frac{\spadesuit}{\heartsuit} \\ y * \left(\frac{\diamondsuit}{\clubsuit} + 1\right) = \frac{\spadesuit}{\heartsuit} \\ y = \frac{\frac{\spadesuit}{\heartsuit}}{1 + \frac{\diamondsuit}{\clubsuit}} \\ y = \frac{\spadesuit \clubsuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)} \end{align}

Now that you have $y$ you can plug into the original equation to find that:

$$ x = \frac{\diamondsuit}{\clubsuit}y = \frac{\diamondsuit}{\clubsuit}*\frac{\spadesuit \clubsuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)} = \frac{\diamondsuit \spadesuit}{\heartsuit\left(\clubsuit + \diamondsuit\right)} $$

I used the symbolds $\diamondsuit$, $\clubsuit$, $\spadesuit$, and $\heartsuit$ to hopefully illustrate that variables are symbols--and you can use any symbols you like.