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?
$x$: 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.
Let one bag of apple weight $x$ pounds and one bag of orange weight $y$ pounds. We then have
\begin{align}
3x&amp;+2y=32\qquad\\
4x&amp;+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
$$
2\cdot8  +  4 =16+4 =20\text{ pounds}.
$$