Cutting cake into 5 equal pieces 
If a cake is cut into $5$ equal pieces, each piece would be $80$ grams
  heavier than when the cake is cut into $7$ equal pieces. How heavy is
  the cake?

How would I solve this problem? Do I have to try to find an algebraic expression for this? $5x = 7y + 400$?
 A: The way to deal with such a question is often to break it down into pieces, and write an equation for each piece.
So suppose the whole cake is $c$ grams (c for cake, name the thing you want to find)
Then one seventh of the cake is $s=\frac c 7$ (s for seven)
And one fifth of the cake is  $ f=\frac c 5$ (f for five)
And you also have the equation $f=s+80$
That deals with all the information you have been given in an orderly fashion - except it involves a lot of extra variable names, and you only want to know about $c$.  You can take any of the three equations and use the others to substitute out $f$ and $s$ to give you a single equation in $c$ alone, which you should be able to solve. In fact the third one is easiest, even though it doesn't involve $c$ at all at the moment. But I'll leave that to you.
I really wanted to point out that it is sometimes easier to write down a number of very simple equations, and solve these, than to try to get directly to a single equation, when it is easy to make a mistake.
A: Cut the cake into $35$ (that is, $5\times 7$) equal pieces. Let the weight of each piece be $p$. Dividing the cake between $5$ people means each person gets $7p$. Dividing between $7$ people means each person gets $5p$. 
So $7p-5p=80$, and therefore $p=40$, and the whole cake weighs $(35)(40)$.
A: Let the weight of cake be $35t$ grams. 
So, the weight of each piece when the cake was cut into $7$ pieces $= \dfrac{35t}{7} = 5t.$
Also, the weight of each piece when the cake was cut into $5$ pieces = $\dfrac{35}{5} = 7t.$
$$7t = 5t + 80$$
$$7t - 5t = 80$$ 
$$2t = 80$$
$$t = 40$$
So, the  weight of the cake is $35 \cdot 40 = 1400$ grams.
A: $$\frac{w}{5}=\frac{w}{7}+80$$
Multiply both sides by 35
$$7w=5w+80\cdot 35$$
substracting both sides by $$2w$$
$$2w=80\cdot 35$$
dividing both sides by 2
$$w=40\cdot 35$$
So weight of cake is 1400
A: If cut it into 5 pieces, but decide you wanted 7 instead, so you cut away 80 grams from each slice, and make two (not so pretty) pieces from that. Now you have 7 pieces of equal size, which means you've got 400g evenly divided among the two ugly out of the seven pieces, so the whole cake weighted 1400g.
A: The natural algebraic approach is to give the unknown weight of the cake a name, $x$, and then translate the information we have into an equation:
$$ \frac{x}{5} = \frac{x}{7}+80 $$
From there it is just algebra: multiply everything by 5 and then 7 to clear denominators, rearrange to separate the multiples of $x$ from the constants, solve.
A: Suppose original weigth of cake is $7x.$ When it is cut into $5$ pieces, each piece weighs $80$ grams more than when it cut into $7$ pieces. So, the equations to model the problem would be
$$5(x+80) = 7x$$
$$5x + 400 = 7x$$
$$x = 200$$
So original weight of cake is $7x = 7 \cdot 200 = 1400$ grams
A: Let $w$ be the weight of the cake in grams.
If you cut the cake into $5$, one slice would weigh $\frac{w}{5}$ grams.  If it was cut into $7$, it would weigh $\frac{w}{7}$ grams.  We know
$$\frac{w}{5}=\frac{w}{7}+80$$
Solve for $w$ (the weight of the cake in grams).
A: The first step is to turn the word problem into an equation; one-fifth of the cake is $80$ grams heavier than one-seventh of the cake, so one-fifth of the cake equals one-seventh of the cake plus 80. "The cake" (specifically its mass) is $x$, and we can work from there:
$$\dfrac{x}{5} = \dfrac{x}{7}+80$$
$$\dfrac{x}{5} - \dfrac{x}{7} = 80$$
Here comes the clever bit; multiply each fraction by a form of one that will give both fractions the same denominator:
$$\dfrac{7}{7}\cdot \dfrac{x}{5} - \dfrac{5}{5}\cdot\dfrac{x}{7} = 80$$
$$\dfrac{7x}{35} - \dfrac{5x}{35} = 80$$
$$\dfrac{2x}{35} = 80$$
$$2x = 2800$$
$$x = 1400$$
You can check your answer by plugging it into the original equation; if the two sides are indeed equal the answer is correct:
$$\dfrac{1400}{5} = \dfrac{1400}{7} + 80$$
$$280 = 200 + 80$$
$$\ \ \ \ \ \ \ \ \ \ \ \ 280 = 280 \ \ \text{<-- yay!}$$
