Helping my daughter with her homework: solving an algebra word problem. Three bags of apples and two bags of oranges weigh $32$ pounds. 
Four bags of apples and three bags of oranges weigh $44$ pounds. 
All bags of apples weigh the same. All bags of oranges weigh the same. 
What is the weight of two bags of apples and one bag of oranges?
 A: Let $x$ be the weight of a bag of apples and $y$ the weight of a bag of oranges. We’re told that $3x+2y=32$ and $4x+3y=44$:
$$\left\{\begin{align*}
&3x+2y=32\\
&4x+3y=44\;.
\end{align*}\right.\tag{1}$$
We want to know what $2x+y$ is.
One way to answer the question is to solve $(1)$ for $x$ and $y$ and substitute into $2x+y$. Multiply the top equation of $(1)$ by $3$ and the bottom by $2$, so as to get equations with the same coefficient on $y$:
$$\left\{\begin{align*}
&9x+6y=96\\
&8x+6y=88\;.
\end{align*}\right.\tag{2}$$
If you now subtract the bottom equation in $(2)$ from the top you find that $x=8$. Substitute that value of $x$ into any of the equations in $(1)$ or $(2)$ to find $y$; I’ll use the top equation in $(1)$, since it has the smallest coefficients. From it I find that $3\cdot8+2y=32$, $24+2y=32$, $2y=8$, and $y=4$. Thus, $2x+y=2\cdot8+4=20$.
If you happen to notice that $2(3x+2y)-(4x+3y)=2x+y$, you can take advantage of a shortcut (which I see Marvis has already pointed out), but if not, solving the system is guaranteed to work, and fairly mechanically, too.
A: Let the weight of $1$ bag of apple be $x$, and the weight of $1$ bag of orange be $y$. Then you have
\begin{equation}
3x + 2y = 32
\end{equation}
and
\begin{equation}
4x + 3y = 44
\end{equation}
Now, solving these equations gives you $x = 8$ and $y = 4$. Thus, the weight of $2$ bags of apple and $1$ bag of orange is $2x+ y = 16 + 4 = 20$.
A: $x$: weight of a bag of apples (in pounds)
$y$: weight of a bag of oranges (in pounds)
First we "translate" the givens into algebraic equations:


*

*$(1)$ "Three bags of apples and two bags of oranges weigh $32$ pounds." $\implies 3x + 2y = 32$.

*$(2)$ "Four bags of apples and three bags of oranges weigh $44$ pounds." $\implies 4x + 3y = 44$


This gives us the system of two equations in two unknowns:
$$3x + 2y = 32\tag{1}$$
$$4x + 3y = 44\tag{2}$$
Ask your daughter to solve the system of two equations in two unknowns to determine the values of $x$ and $y$. 
Hints for your daughter: 


*

*multiply equation $(1)$ by $3$, and multiply equation $(2)$ by $2$: 


$$9x + 6y = 96\tag{1.1}$$
$$8x + 6y = 88\tag{2.1}$$


*

*subtract equation $(2.1)$ from equation $(1.1)$, which will give the value of $x$. 

*Solve for $y$ using either equation $(1)$ or $(2)$ and your value for $x$.

*Then determine what $2x + y$ equals. That will be your (her) solution.
A: Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have
\begin{align}
3x+2y & = 32\\
4x+3y & = 44
\end{align}
We need the weight of $2$ bags of apples and $1$ bag of orange i.e. we need $2x+y$.
Note that $$2x+y = 2(3x+2y) - (4x+3y) = 2 \times 32 - 44 = 20$$
A: Here's another approach that shows how one might be able to intuit the answer for this particular problem without being able to solve general systems of linear equations:
"We know that 3 bags of apples and 2 bags of oranges weigh a total of 32 pounds, and we know that 4 bags of apples and 3 bags of oranges weigh 44 pounds.  Then by subtracting the first from the second, we find out that one bag of apples and one bag of oranges weighs 44-32=12 pounds.
But now that we have this, we can take a bag of apples and a bag of oranges away from the 3-and-2-bags pile to see that 2 bags of apples and 1 bag of oranges weighs 32-12=20 pounds."
(This is, of course, roughly equivalent to Marvis's observation, but involves no multiplication whatsoever, just a couple of successive subtractions.  And you could go on to note that you can take another bag-of-apples-and-bag-of-oranges away from the 2-and-1 pile to find that a bag of apples weighs 20-12=8 pounds all by itself and then that a bag of oranges weighs 12-8=4 pounds, but that's tangential and of course mostly a function of this special set of parameters.)
A: Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have
\begin{align}
3x&+2y=32\qquad\\
4x&+3y=44\qquad
\end{align}
To find the value of $x$ and $y$:
\begin{align}
(3x+2y=32)  \cdot 3\\
(4x+3y=44)  \cdot 2
\end{align}
yields
\begin{align}
9x+6y=96 \qquad(1)\\
8x+6y=88 \qquad(2)
\end{align}
Subtracting $(2)$ from $(1)$, we get $x=8$ and $y = 4$. So the weight of $2$ bags of apples and $1$ bag of orange is
$$
2\cdot8  +  4 =16+4 =20\text{ pounds}.
$$
