Lesson Solving word problems using linear systems of two equations in two unknowns

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<H2>Solving word problems using linear systems of two equations in two unknowns</H2>

In this lesson we present some typical word problems and show how to solve them using linear systems of two equations in two unknowns.

<H3>Problem 1. The Madison Local High School band</H3>The Madison Local High School marching band sold gift wrap to earn money for a band trip to Orlando, &nbsp;Florida. 
The gift wrap in solid colors sold for &nbsp;$4.00&nbsp; per roll, &nbsp;and the print gift wrap sold for &nbsp;$6.00&nbsp; per roll. 
The total number of rolls sold was &nbsp;480, &nbsp;and the total amount of money collected was &nbsp;$2340. 
How many rolls of each kind of gift wrap were sold?

<B>Solution</B>

Let &nbsp;<B>x</B>&nbsp; be the unknown number of gift wrap in solid colors and &nbsp;<B>y</B>&nbsp; be the unknown number of print gift wraps. 
Since the total number of rolls sold was &nbsp;480, &nbsp;the first equation is
{{{x + y = 480}}}.
Now, &nbsp;the total cost of &nbsp;<B>x</B>&nbsp; gift wraps in solid colors sold for &nbsp;$4.00&nbsp; per roll is equal to &nbsp;{{{4*x}}} dollars, 
while the total cost of &nbsp;<B>y</B>&nbsp; print gift sold for &nbsp;$6.00&nbsp; per roll is equal to &nbsp;{{{6*y}}}&nbsp; dollars. 
Since the total amount of money collected was &nbsp;$2340, &nbsp;the second equation is
{{{4x + 6y = 2340}}}.

So, &nbsp;we reduced our problem to the solution of the linear system of two equations in two unknowns

{{{system(x +  y = 480,
         4x + 6y = 2340
)}}}.

Let us &nbsp;<B><U>apply the substitution method</U></B>&nbsp; (see the lesson &nbsp;<A HREF= http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-system-of-two-eqns-by-the-Subst-method.lesson> Solution of the linear system of two equations in two unknowns by the Substitution method</A>).

Express &nbsp;{{{y}}}&nbsp; from the first equation:
{{{y = 480-x}}}.
Substitute this to the second equation:
{{{4*x + 6*(480-x) = 2340}}}.
Open parentheses, &nbsp;collect common terms, &nbsp;and step-by-step simplify:
{{{4X + 6*480 -6x = 2340}}},
{{{-2x + 2880 = 2340}}},
{{{-2x = 2340-2880}}},
{{{2x = 540}}}
{{{x = 270}}}.
Substitute the found value of &nbsp;{{{x=270}}}&nbsp; to the first equation and calculate &nbsp;{{{y}}}:
{{{270 + y = 480}}},
{{{y = 480 - 270 = 210}}}.

As a result, &nbsp;you get &nbsp;{{{x=270}}}, &nbsp;{{{y=210}}}&nbsp; as the potential solution.

<B>Check</B> 
Substitute these values of &nbsp;{{{x=270}}}&nbsp; and &nbsp;{{{y=210}}}&nbsp; into the first and the second equations.
You have 
{{{x+y = 270+210 = 480}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;for the left side of the first equation, &nbsp;and
{{{4x+6y = 4*270+6*210 = 1080 + 1260 = 2340}}} &nbsp;&nbsp;&nbsp;&nbsp;for the left side of the second equation.

<B><U>Answer</U></B>. &nbsp;1080 gift wrap in solid colors and &nbsp;1260&nbsp; print gift were sold.

<H3>Problem 2. Apples and oranges</H3>If &nbsp;4&nbsp; apples and &nbsp;2&nbsp; oranges cost &nbsp;$1&nbsp; and &nbsp;2&nbsp; apples and &nbsp;3&nbsp; orange cost &nbsp;$0.70, &nbsp;how much does each apple and each orange cost? 
There are no quantity discounts.

<B>Solution</B>

Let &nbsp;{{{x}}}&nbsp; be the unknown price for one apple as cents and &nbsp;{{{y}}}&nbsp; be the unknown price for one orange as cents.
From the first condition you have the equation
{{{4x + 2y = 100}}},
while from the second problem condition you have another equation
{{{2x + 3y = 70}}}.

So, &nbsp;our problem is reduced to the solution of the linear system of two equations in two unknowns

{{{system(4x + 2y = 100,
          2x + 3y =  70
)}}}.

Let us &nbsp;<B><U>apply the elimination method</U></B>, &nbsp;which is &nbsp;<B>multiplication and addition/subtraction method</B>
(see the lesson &nbsp;<A HREF= http://www.algebra.com/algebra/homework/coordinate/lessons/Solution-of-the-lin-syst-of-two-eqns-with-two-unknowns-Elimination-method.lesson>Solution of the linear system of two equations in two unknowns by the Elimination method</A>).

Keep the first equation as is and multiply the second equation by two:

{{{system(4x + 2y = 100,
          4x + 6y = 140
)}}},

Subtract the second equation from the first one:

{{{0x - 4y = -40}}},
{{{y = 10}}}.

Now, &nbsp;substitute the found value of &nbsp;{{{y = 10}}}&nbsp; into the first equation and calculate &nbsp;{{{x}}}:
{{{4x + 2*10 = 100}}},
{{{4x + 20 = 100}}},
{{{4x = 100 - 20 = 80}}},
{{{x = 20}}}.

So, &nbsp;you get &nbsp;{{{x=20}}}, &nbsp;{{{y=10}}}&nbsp; as the potential solution.

<B>Check</B> 
Substitute these values of &nbsp;{{{x=20}}}&nbsp; and &nbsp;{{{y=10}}}&nbsp; into the first and the second equations.
You have 
{{{4x + 2y = 4*20 + 2*10 = 80 + 20 = 100}}} &nbsp;&nbsp;for the first equation left side, &nbsp;and
{{{2x + 3y = 2*20 + 3*10 = 40 + 30 =  70}}} &nbsp;&nbsp;&nbsp;&nbsp;for the second equation left side.

<B><U>Answer</U></B>. &nbsp;Each apple costs &nbsp;20 cents&nbsp; and each orange costs &nbsp;10 cents.

<H3>Problem 3. Tickets</H3>Tickets are sold at &nbsp;$4.00 &nbsp;for adults and &nbsp;$2.50&nbsp; for students. &nbsp;If 100 tickets were sold for &nbsp;$355.00, 
how many tickets were adult tickets?

<B>Solution</B>

Let &nbsp;"x"&nbsp; be the numbers of adult tickets, &nbsp;and
let &nbsp;"y"&nbsp; be the numbers of student tickets.

The system is:

&nbsp;&nbsp;x + &nbsp;&nbsp;&nbsp;&nbsp;y = 100, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;("100 tickets were sold") &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and
4x + 2.5y = 355. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;("100 tickets were sold for $355.00")

To solve the system, &nbsp;express &nbsp;"x"&nbsp; from &nbsp;(1): &nbsp;&nbsp;x = 100 - y, &nbsp;and substitute it into &nbsp;(2). &nbsp;You will get a single equation for y:

4*(100 - y) + 2.5y = 355.

Simplify and solve it:

400 - 4y + 2.5y = 355, &nbsp;&nbsp;&nbsp;&nbsp;or

-1.5y = 355 - 400,

-1.5y = -45,

y = {{{(-45)/(-1.5)}}} = 30.

So, &nbsp;we just found y, &nbsp;the number of student tickets.

Then  &nbsp;x = 100 - y = 100 - 30 = 70. &nbsp;It is the number of adult tickets.

<B>Check</B>: &nbsp;&nbsp;70*4 + 30*2.5 = 280 + 75 = 355.  &nbsp;&nbsp;&nbsp;&nbsp;Correct !

<B>Answer</B>. &nbsp;&nbsp;70 &nbsp;adult and &nbsp;30 &nbsp;student tickets were sold.

<H3>Problem 4. Alloys</H3>A piece of metal is &nbsp;1 ft. long, &nbsp;6 in. wide, &nbsp;and &nbsp;4 in. thick, &nbsp;and weighs &nbsp;189.8125 pounds.  
It is composed of an alloy of gold and copper.  &nbsp;Determine percentage of gold.

Gold density is 0.70 lbs. per cu. in.

Copper density is 0.32 lbs. per cu. in.

<B>Solution</B>

When they ask &nbsp;"Determine percentage of gold", &nbsp;they mean the ratio of the gold contents by &nbsp;<U>WEIGHT</U> &nbsp;to the total &nbsp;<U>WEIGHT</U> &nbsp;of the alloy; 
the ratio, expressed as percentage.

<pre>
The volume of a piece of metal is  (12 in x 6 in x 4 in) = 288 cubic inches.

The density of the alloy is  {{{189.8125/288}}} = 0.659 lb/in^3.

Now, having the densities expressed in consistent units, you can write the equation.

Let  0 <= x <= 1 be the fraction (by the weight) of gold in the alloy (i.e. the percentage is 100x).

So the weight fraction of copper is (1-x), obviously.

Then the "fraction" equation is

{{{D[gold]*x}}} + {{{D[copper]*(1-x)}}} = {{{D[alloy]}}},

where D stands for density, or

0.70*x + 0.32*(1-x) = 0.659.

Simplify and solve for x:

0.70x + 0.32 - 0.32x = 0.659  ====>  0.38x = 0.659 - 0.32 = 0.339  ====>  x = {{{0.339/0.38}}} = 0.8921.

Thus the gold fraction by the weight is  0.8921,  or  89.21%.
</pre>

My other lessons on solving linear systems of two equations in two unknowns in this site are

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