Problems

Problem means a situation, matter, or person that presents perplexity or difficulty.Also a statement requiring a solution usually by means of one or more operations or geometric constructions.

Problem Solving Guidelines

1. Understand the Problem
Read the problem carefully and decide which quantities are unknown.

2. Develop a Plan
Represent one of the unknown values by our variable and the second unknown by a second variable.

3. Carry out the Plan
Study the facts until you understand their meaning. Then translate the related facts into equation in two variables.

4. Looking Back
Check answers directly against the facts of the problem not the equations.Write a statement to answer the equation being asked in the problem.


A. Number Word Problem


The sum of twice a number and 5 is 9. What is the number?

How to solve it

Use the following relationship to set up an equation:

[ twice the number ] + 5 = 9

Solution

Step 1: Let x represent the number.

Step 2: Write an algebraic expression for "twice the number" in terms of x.

twice the number is 2x.

Step 3: Use the relationship to set up an equation.

2x + 5 = 9

Step 4:

Solve the equation for x.

2x + 5 = 9
2x = 4
x = 2

So the number is 2.


Problem No.2

Four times a number decreased by 10 is equal to the sum of twice the number and 2. find the number.

How to solve it:

Use the following relationship to set up an equation:

[four times the number decreased by 10] = [the sum of twice the number and 2]

Solution:

Step 1: Let x represent the number.

Step 2: Write an algebraic expression for "four times the number decreased by 10" in terms of x.

four times the number decreased by 10 = 4x - 10.

Write an algebraic expression for "the sum of twice the number and 2" in terms of x.
the sum of twice the number and 2 = 2x + 2.


Step 3: Use the relationship to set up an equation.

4x - 10 = 2x + 2

Step 4:

Solve the equation for x.

4x - 10 = 2x + 2
2x = 12
x = 6

So the number is 6.


Geometry Word Problems Involving Perimeter

Example 1:

A triangle has a perimeter of 50. If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?

Solution:

Step 1:: Assign variables:

Let x = length of the equal side


Step 2: Write out the formula for perimeter of triangle.

P = sum of the three sides

Step 3: Plug in the values from the question and from the sketch.

50 = x + x + x+ 5

Combine like terms

50 = 3x + 5

Isolate variable x

3x = 50 – 5
3x = 45
x =15

Be careful! The question requires the length of the third side.
The length of third side = 15 + 5 =20

Answer: The length of third side is 20


Example No.2

In a quadrilateral two angles are equal. The third angle is equal to the sum of the two equal angles. The fourth angle is 60° less than twice the sum of the other three angles. Find the measures of the angles in the quadrilateral.

Solution:

Step 1: Assign variables:

Let x = size of one of the two equal angles

Step 2: Write down the sum of angles in quadrilateral.

The sum of angles in a quadrilateral is 360°

Step 3: Plug in the values from the question and from the sketch.

360 = x + x + (x + x) + 2(x + x + x + x) – 60
Combine like terms

360 = 4x + 2(4x) – 60
360 = 4x + 8x – 60
360 = 12x – 60

Isolate variable x

12x = 420
x = 35

The question requires the values of all the angles.
Substituting x for 35, you will get: 35, 35, 70, 220

Answer: The values of the angles are 35°, 35°, 70° and 220°

Reference:
http://www.idealmath.com/algebra/equationone/numprob01.htm
http://www.onlinemathlearning.com/geometry-word-problems.html

A System Equations Solved by Algebraic Methods

Solution by Substitution

This method is easily applied when at least one equation gives the value of one unknown of the other.

To solve a system of Linear Equation in two variables:

1. Use either of the equations to solve for one variables.(Always select the equation where in the variable is easier to work with).

2. Substitute the expression obtained in step 1 in the other equation.

3. Solve the resulting equation to find the value of the variable.

4. Substitute the value of this variables in the simpler equation to find the value of the other variable.

5. The resulting no pair is the solution.

6. Check the results in the two original equations.

Examples:
A. Solve the system:
3x + 2y = 11 (1)
y = 4x (2)

Solution:
3x + 2y = 11
y = 4x

Substitute 4x for y in equation (1)

3x + 2(4x) = 11
3x + 8x = 11
11x = 11
x = 11/11 or 1

Substitute 1 for x in equation (2)

y = 4(1)
y = 4

Check:
3x + 2y = 11
3(1)+ 2(4) = 11
11 = 11 /

y = 4x
4 = 4(1)
4 = 4 /

The solution is the ordered pair (1,4).

B. Solve the system:
5x + 3y = 3 (1)
y = 8 - 4x (2)


Solution:
5x + 3y = 3
y = 8 - 4x

Substitute 8 - 4x for y in equation (1)

5x + 3 (8 - 4x) = 3
5x + 24 - 12x = 3
-7x = -21
x = 3


Substitute 3 for x in equation (2)

y = 8 - 4(3)
y = -4


Check:
5x + 3y = 3
5(3)+ 3(-4)= 3
3 = 3 /


y = 8 - 4x
-4 = 8 - 4(3)
-4 = -4 /

The solution is the ordered pair (3,-4).


Solution by Elimination

This method of solving systems of equations depends on the addition property of equality, which states that when two quantities are equal, adding the same number to each quantity results in equal sums.This is used when coefficient of one of the unknown quantities in the two equations are equal in absolute value.

To solve a system of linear equation in two variable:

1. Arrange the two given equations in standard form.

2. Reduce the two equations in two variables to a single equation in one variable.

a. If both equations have a variable with the same numerical coefficients, eliminate it by:

(1) Adding the two equations if there coefficients have the same opposite signs.
(2) Subtracting the two equations if there coefficients have the same sign.

b. When neither variable has the same numerical coefficients multiply one or both of the given equations by numbers which will make the numerical coefficients of one if the variables the same in both equations and proceed as in (a).

3. Solve the resulting equation for the value of the remaining variable.

4. Substitute this value in one of the given equations, to find the value of the other variable.

5. The resulting number pair is the solution.

6. Check the values in two given equations.


A. Solve the system:
x + y = 8 (1)
x - y = 2 (2)

Solution:
(x + y)+(x - y) = 8 + 2
2x = 10
x = 5

Take equation (1),
x + y = 8
5 + y = 8
y = 3

Check:
Substitute the ordered pair in both equations.
x + y = 8
5 + 3 = 8
8 = 8 /


x - y = 2
5 - 3 = 2
2 = 2 /

The solution is the ordered pair (5,3).


Solution by Comparison

This method of solving of systems of linear equations uses the transitive property of equality, which states that quantities equal to the same quantity are equal. This is used when the same unknown in the two equations has coefficient 1. The value of the unknown can then be found.

A. Solve the system:
x + 2y = 4 (1)
x - 3y = -1 (2)

Solution:

a. Solve both equations for the variable x to find y.
x = -2y + 4 (1)
x = 3y - 1 (2)

Equate equation (1) and (2) since both are equal to x.
-2y + 4 = 3y - 1
-5y = -5
y = 1

b. Solve both equations for the variable y to find x.
2y = -x 4 (3)
y = -1/2x + 2

-3y = -x - 1 (4)
y = 1/3x + 1/3

Equate equations (3) and equation (4) since both are equal to y.
-1/2x +2 = 1/3x + 1/3

Multiply both sides of the equation by 6.
6 (-1/2x +2) = (1/3x + 1/3)6
-3x + 12 = 2x + 2
x = 2

Check:
x + 2y = 4
(2) + 2(1) = 4
4 = 4 /

x - 3y = -1
(2) - 3(1) = -1
-1 = -1 /

The solution is the ordered pair (2,1).


Reference:
Intermediate Algebra
Authors: Soledad Jose-Dilao Ed.D.
Julieta G. Bernabe