Intermediate Algebra
Tutorial 21: Systems of Linear Equations and Problem Solving
Learning Objectives
Introduction
Hey, lucky you, we have another tutorial on word problems. As mentioned before, whether you like it or not, whether you are going to be a mother, father, teacher, computer programmer, scientist, researcher, business owner, coach, mathematician, manager, doctor, lawyer, banker (the list can go on and on), problem solving is everywhere. Some people think that you either can do it or you can't. Contrary to that belief, it can be a learned trade. Even the best athletes and musicians had some coaching along the way and lots of practice - that's what it also takes to be good at problem solving.
The word problems in this section all involve setting up a system of linear equations to help solve the problem. Basically, we are combining the concepts from Tutorial 8: An Introduction to Problem Solving, Tutorial 19: Solving Systems of Linear Equations in Two Variables and Tutorial 20: Solving Systems of Linear Equations in Three Variables all rolled up into one tutorial. We will be looking at different types of word problems involving such ideas as distance, percentages, and something we can all relate to MONEY!!!
Tutorial
In the problems on this page, we will be letting each unknown be a separate variable. So, if you have two unknowns, you will have two variables, x and y. If you have three unknowns, you will have three variables, x, y, and z.
In the problems on this page, we will be setting up systems of linear equations. The number of equations need to match the number of unknowns. For example, if you have two variables, then you will need two equations. If you have three variables, then you will need three equations.
Since we are looking for three numbers, we will
let
x = the smallest number
y = middle number
z = the largest number
Equation (1):
Equation (2):
Equation (3):
Simplify and put all three equations in the form Ax + By + Cz = D if needed.
Equation (2) needs to be put in the correct form:
Since y is already eliminated in equation (4) and (3), it would be quickest and easiest to eliminate y.
We can use equation (4) as one equation with y eliminated:
We can use equation (3) as another equation with y eliminated:
Putting those two equations together we get:
Multiplying equation (4) by -2 and then adding that to equation (3) we get:
*z's have opposite coefficients
*z's dropped out
*Inverse of sub. 8 is add 8
Using equation (1) to plug in 2 for x and 8 for z and solving for y we get:
*Inverse of add 10 is sub. 10
Final Answer:
2 is the smallest number, 4 is the middle number and 8
is the largest
number.
Since we are looking for two different amounts, we will
let
x = the number of gallons of 20% alcohol solution
y = the number of gallons of 50% alcohol solution
Equation (1):
Equation (2):
Simplify if needed.
We can simplify this by multiplying both sides of equation (2) by 10 and getting rid of the decimals:
Multiply one or both equations by a number that will create opposite coefficients for either x or y if needed AND add the equations.
If we multiply equation (1) by -2, then the x’s will have opposite coefficients.
Multiplying equations (1) by -2 and then adding that to equation (3) we get:
*x's
have opposite
coefficients
*x's dropped out
Solving for y we get:
Using equation (1) to plug in 3 for y and solving for x we get:
*Inverse of add 3 is sub. 3
Final Answer:
6 gallons of 20% solution and 3 gallons of 50% solution
Since we are looking for two different rates, we will
let
x = rate of the plane
y = the rate of the wind
Since this is a rate/distance problem, it might be good to organize our information using the distance formula.
Keep in mind that the wind speed is affecting the overall speed.
When the plane is with the wind, it will be going faster. That rate will be x + y.
When the plane is going against the wind, it will be
going slower.
That rate will be x - y.
Equation (1):
Equation (2):
Simplify if needed.
We can simplify this by dividing both sides of equation (1) by 2 and equation (2) by 2.5 getting rid of the ( ) and decimals at the same time:
*Div. both sides of eq. (2) by
2.5
Multiply one or both equations by a number that will create opposite coefficients for either x or y if needed AND add the equations.
Since we already have opposite coefficients on y, we can go right into adding equations (3) and (4) together:
*y's dropped out
Solving for x we get:
Using equation (3) to plug in 270 for x and solving for y we get:
Final Answer:
The airplane speed is 270 mph and the air speed is 30 mph
C(x) = 20x +
50000
R(x) = 25x
We will let,
x = the number of units
C(x) = 20x + 50000
R(x) = 25x
This problem appears a little different because of the
function notation.
Keep in mind that function notation translates to being y.
Cost function:
C(x) = 20x + 50000
Revenue function:
R(x) = 25x
Since the break-even point is when revenue = cost, we will go right into setting this up using the substitution method as discussed in Tutorial 19: Solving Systems of Linear Equations in Two Variables.
*Inverse of mult. by 5 is div. by 5
Final Answer:
10000 units are needed to break-even
Practice Problems
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1c: Solve the word problem.
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Last revised on July 10, 2011 by Kim Seward.
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