Writing Equations for Word Problems

Sometimes figuring out how to write an equation for solving a word problems is the hardest part of the solution.

Here are some examples of word problems and the equations they lead to:

The sum of two numbers is 90. The difference between them is 75.
Find the two numbers.

Let a and b stand for the two numbers.

What do you know?

You are adding and subtracting the same two numbers.

Can you write one equations using addition and one using subtraction?

(Sum) = a + b = 90

(Difference) = a - b = 75

You could solve these equations using substitution or elimination.



A wildlife management station cares for sick birds and deer. The animals being cared for have a total of 25 heads and 74 feet (no heads or feet are missing!). How many of each animal are there?

Let b = birds and d = deer

What do we know?

If there are 25 heads, there must be a total of 25 animals

b + d = 25


What else do we know? Birds have two feet each and deer have four feet each and all the feet total 74.

2b + 4d = 74

Now we could solve by either substitution or elimination.


Breakfast: If two eggs with bacon cost £2.70 and one egg with bacon costs £1.80, what does bacon cost alone?

What do we know?

If we use e for eggs and b for bacon, we can write equations for each meal:

two eggs and bacon costs 2.70

2e + b = 2.7

one egg with bacon costs 1.80

e + b = 1.8

We can use substitution or elimination to solve these equation.

Walter is riding his bike to Thames City which is 28 miles away. It takes him one hour to make the ride against a head wind but it only takes him 48 minutes to return to St. John's Wood with a tail wind.

How fast was he riding without the wind and how strong was the wind?

If a is his average speed and w is the speed of the wind..... what else do we know?

(a - w) would be his speed (rate) riding against the wind.

(a + w) would be his speed (rate) riding with the wind

The distance for both parts of the trip is 28 miles.

It took one hour on the ride against the wind and 48 minutes (48/60 = 4/5 hour) to ride with the wind.

Using d = rt

28 = (a - w)1 (against the wind)

28 = (a + w)(4/5) (with the wind)

Simplify:

(1)28 = (a - w)1

28 = a - w

(5/4)28 = (a + w)(4/5)(5/4)

75 = a + w

We could then use substitution or elimination to solve the problem.

Stephanie received the results of her ERB scores in math and reading. Her reading score is 70 points less than her math score. Her total for the two parts is 1250.

Let m = math score and r = reading score.

What do we know?

Her math score plus her reading score is 1250.

m + r = 1250

What else do we know?

Her reading score is 70 points less than her math score.

r = m - 70

We could use substitution or elimination to solve this problem.

Simultaneous Equations

Simultaneous equations - two or more equations that have one or more points (solutions) in common.


There are three ways to solve simultaneous equations: graphing, substitution and elimination.

Graphing:

To solve for a common solution to two equations using graphing, you graph each line either on paper or on a graphing calculator.

The point at which they intersect is the solution to both equations.

Example: Graph the solution for y = x + 1 and y = (-1/2)x + 4


The solution is the point at which the lines cross.


Some lines have more than one solution.

y = 2x - 2 and 3y = 6x - 6

Lines with the same slope and same y-intercept will be graphed one on top of the other so that all points coincide. All points on the lines are solutions (infinite solutions).



Lines with the same slope but different y-intercepts are parallel and have no points in common. In this situation there is no solution.


Substitution:

To use substitution you must have at least one equation in simplest form that is equal to one of the variables.

Example:
x + 2y = 8 and y = x + 1

Since the second equation is in simplest form (y =), we will use it to substitute.

If y = x + 1 we can substitute the value for y into the other equation:

x + 2(x + 1) = 8

Simplify using distribution:

x + 2x + 2 = 8

Combine like terms:

3x + 2 = 8

Simplify:

3x + 2 - 2 = 8 - 2

3x = 6

Divide both sides by 3.

x = 2

Now we need to solve for y too!

Use either of the original equations: x + 2y= 8 and y = x + 1

y = 2 + 1

y = 3

The solutions to this system of equations is ( 2, 3)




Elimination:

In this form we will eliminate one of the variable to solve for one at a time, similar to substitution but using addition and subtraction.

We will start with the same equations we used in the first two forms but written in Standard Form:

y = x + 1 becomes -x + y = 1

y = -1/2x +4 becomes 2y = -x = 8 when we clear the fraction,

then x + 2y = 8 in Standard Form.

Line the two equations up so that like terms are in a column:

-x + y = 1
x + 2y = 8

Combine like terms vertically.

-x and x cancel each other out eliminating x from the equation.

y + 2y = 3y

1 + 8 = 9

3y = 9

y = 3

Substitute the value for y back into either original equations:

-x + y = 1 or x + 2y = 8

x + 2(3) = 8

x + 6 = 8

x + 6 - 6 = 8 - 6

x = 2

The solution for these two equations is (2, 3).



For more complicated equations we may need to multiply one of the equations to make the coefficient inverse.

2x + 3y = 11
2x + y = 3

We can eliminate the x by multiplying the second equation by -1

2x + 3y = 11
-2x - y = - 3

The x is eliminated leaving:

3y = 11
- y = - 3

Combining those terms gives us

2y = 8

y = 4

Substituting y back in either of the original equations:

2x + 3(4) = 11

2x + 12 = 11

2x + 12 - 12 = 11 - 12

2x = -1

x = - 1/2

Solution (-1/2, 4)


Some equations are even more complicated and you must multiply to get one variable ready for elimination.

2x + 3y = 7
3x - y = 5

Multiply the second equation by 3 to eliminate the y.

2x + 3y = 7
9x -3y = 15

y is eliminated leaving

11x = 22

x = 2

Solve for y.

2(2) + 3y = 7

4 + 3y = 7

4 - 4 + 3y = 7 - 4

3y = 3

y = 1

Solution (2, 1)


For even more complicated equations you may need to multiply both equations.
5x + 2y = 1
2x - 3y = -11

We can multiply the first equation by 2 and the second equation by -5.

2(5x + 2y = 1)
-5(2x - 3y = -11)

10x + 4y = 2
-10x + 15y = 55

The x is eliminated leaving:

4y = 2
15y = 55

19y = 57

divide by 19

y = 3

Substitute the value for y back in either original equation.

5x + 2y = 1

5x + 2(3) = 1

5x + 6 = 1

5x + 6 - 6 = 1 - 6

5x = -5

x = -1

Solution (-1, 3)

Solving Inequalities

In Lessons 4-2, 4-3 and 4-4 we have learned about graphing inequalities, solving when addition and subtraction are involved, solving when multiplication is involved and when there is a mix of operations on one or both sides of the inequality sign.

We always want to get our final answer simplified into the form of:

x > 3

When we are solving for a variable we work backwards through the order of operations:

Subtraction or Addition first

Division or Multiplication second

Clear any addition or subtraction first.

X + 4 < 12

subtract 4 from both sides

X + 4 - 4 < 12 - 4


X < 8


Clear any multiplication or division.


3x > 21


Multiply by the reciprocal

1/3 • 3x > 21 • 1/3


x > 7


For more complex problems such as

2(x - 3) < 4(x - 5)

Use the distributive property to clear the parenthesis first.

2x - 6 < 4x - 20

Then combine like terms by moving the terms with the variable to one side of the inequality and the terms without the variable to the other.

2x - 2x - 6 < 4x - 2x - 20

-6 < 2x - 20



-6 + 20 < 2x - 20 + 20

14 < 2x

Then multiply by the reciprocal.

1/2 • 14 < 2x • 1/2

7 < x or x > 7


Remember to always simplify each side of the inequality before you begin combining like terms.

Graphing Inequalities on a Number Line

When graphing inequalities on a number line there are conventions for the direction of the line and the end point of the line.

The end point is closed (filled in) when the symbols are

≤ less than or equal to
or
≥ greater than or equal

The end point is open when the symbols are

< less than
or
> greater than

The direction of the line on the graph is determined by whether the amount is greater than (going to the right on the number line) or less than (going to the left on the number line).

If the inequality is written with the variable on the left then the inequality symbol points in the direction of the line.

Semester Exam

Congratulations to everyone for studying very diligently for the semester exam. We had our usual assortment of grades: A's, B's C's and for the first time this year, grades lower than C.

I know some of you are disappointed with your results and I have options for you that may help. They will not help change your grade for this semester but they might help you do better in the second semester. Remember you need at 85% for the year to move on to geometry in grade 9.

This blog is one of the things I am going to try to see if it helps you improve not just your grade but your UNDERSTANDING of the algebra concepts. I am not interested in your grade - although you and your parents may be. I am interested in you understanding our algebra concepts SO WELL that you can use them to work any problem.

Toward that end, I will be blogging about our math lessons and send you a link so you can read and think about the concepts.

I hope this is one more thing that will make a difference in your understanding of algebra!