Many word problems, upon translation, result in two equations involving two variables (two ‘unknowns’).
In mathematics, a collection of more than one equation being studied together is called a system of equations.
The systems in this section are fairly simple,
and can be solved by substituting information from one equation
into the other.
The procedure is illustrated in the following example:
Antonio loves to go to the movies. He goes both at night and during the day. The cost of a matinee is \$6.00. The cost of an evening show is \$8.00. If Antonio went to see a total of [beautiful math coming... please be patient] $\,12\,$ movies and spent \$86.00, how many night movies did he attend? 
ENGLISH WORDS  TRANSLATION INTO MATH  NOTES/CONVENTIONS 
“Antonio went to see a total of 12 movies”  [beautiful math coming... please be patient] $n+d = 12$ 
NOTE: There are many realnumber choices for $\,n\,$
and $\,d\,$ that make this equation true. Here are a few:
[beautiful math coming... please be patient]
$0 + 12 = 12$
[beautiful math coming... please be patient]
$1 + 11 = 12$
[beautiful math coming... please be patient]
$1.3 + 10.7 = 12$
[beautiful math coming... please be patient]
$(2) + 14 = 12$
Of course, we want whole number solutions, and we also need something else to be true.

“... and spent \$86.00”  [beautiful math coming... please be patient] $8n + 6d = 86$ 
Each night movie costs \$8.00, so $\,n\,$ night movies cost $\,8n\,$ dollars. Each day movie costs \$6.00, so $\,d\,$ day movies cost $\,6d\,$ dollars. Both [beautiful math coming... please be patient] $\,8n\,$ and $\,6d\,$ have units of dollars. Also, the number $\,86\,$ has units of dollars. It's important that you have the same units on both sides of the equal sign. Here, we have: dollars plus dollars is dollars. CONVENTION: Write [beautiful math coming... please be patient] $\,8n\,$, not (say) $\,8.00n\,$ or $\,\$8n\,$ or $\,\$8.00n\,$. NOTE: Convince yourself that there are also infinitely many realnumber choices for [beautiful math coming... please be patient] $\,n\,$ and $\,d\,$ that make this equation true. We want a choice for [beautiful math coming... please be patient] $\,n\,$ and a choice for $\,d\,$ that make BOTH equations true at the same time. 
[beautiful math coming... please be patient] $8(12  d) + 6d = 86$  original equation 
[beautiful math coming... please be patient] $96  8d + 6d = 86$  distributive law 
[beautiful math coming... please be patient] $96  2d = 86$  combine like terms 
[beautiful math coming... please be patient] $2d = 10$  subtract $\,96\,$ from both sides 
[beautiful math coming... please be patient] $d = 5$  divide both sides by$\,2\,$ 
[beautiful math coming... please be patient] $n+d = 12$  the simple equation 
[beautiful math coming... please be patient] $n + 5 = 12$  substitute in the known information 
[beautiful math coming... please be patient] $n = 7$  subtract $\,5\,$ from both sides 
EQUATIONS  CHECK  TRUE? 
[beautiful math coming... please be patient] $n + d = 12$  [beautiful math coming... please be patient] $7 + 5 \,\,\overset{\text{?}}{ = }\,\, 12$  Yes! 
[beautiful math coming... please be patient] $8n + 6d = 86$  [beautiful math coming... please be patient] $8(7) + 6(5) \,\,\overset{\text{?}}{ = }\,\, 86$ 
Yes! (Feel free to use your calculator.) 
Antonio loves to go to the movies. He goes both at night and during the day. The cost of a matinee is \$6.00. The cost of an evening show is \$8.00. If Antonio went to see a total of $\,12\,$ movies and spent \$86.00, how many night movies did he attend? 
Let
[beautiful math coming... please be patient]
$\,n = \text{# night tickets}\,$. Let [beautiful math coming... please be patient] $\,d = \text{# day tickets}\,$.
[beautiful math coming... please be patient] $8(7) + 6(5) \,\,\overset{\text{?}}{=}\,\,86$ ☺ Antonio attended 7 night movies. 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
