Equations and Inequalities II - Weebly
Unit 1 Graphing a System of Equations
An equation is a sentence formed by two mathematical expressions containing one or more variables and related by the = sign.
A system of equations of the first degree is a set of equations which must be solved simultaneously.
An ordered pair is a solution-pair of an equation of the first degree in two variables if, after
substitution, the ordered pair satisfies the equation or expresses a true statement.
The graphic solution of a system of equations consists in determining the coordinates of the point that is common to the two lines of the system, if there is a point of intersection.
To solve a system of equations of the first degree in two variables graphically:
1. Transform each of the equations in the system to the form y = m x + b.
2. Determine at least three solution-pairs for each equation and complete a table of values.
3. Graph each equation in the same Cartesian plane.
4. Identify the point of intersection of the two lines , if there is one.
5. Determine the solution of the system of equations.
In solving a system of equations of the first degree in two variables graphically, three situations can occur:
1. The two lines can intersect and , in this case , the solution of the system is a unique solution-pair.
2. The two lines can be parallel and distinct and, in this case, there is no solution.
3. The two lines can coincide and, in this case, there would be an infinite number of solutions.
Unit 2 Solving a System of Equations by Comparison
Solving a system of equations of the first degree in two variables means determining all the ordered pairs
( x , y ) which satisfy the equations when substituted for the variables.
To solve a system of equations of the first degree in two variables by comparison:
1. Isolate the same variable in both equations.
2. Compare the two equations obtained.
3. Solve this new equation in one variable.
4. Substitute the value obtained in either one of the equations.
5. Verify the solution in each of the original equations of the system.
6. State the solution.
A system of parallel and distinct lines does not have a common solution and it leads to an impossible situation in terms of an (1) algebraic solution or (2) there is no algebraic solution possible.
In any system of equations of the form: (1). y = m1 x + b1
(2). y = m2 x + b2
If m1 = m2 and b1 ≠ b2, then the two lines are parallel and distinct , and the system does not have a solution.
A system of parallel and coinciding lines has an infinite number of solutions and leads to an indeterminate solution when solves algebraically.
In any system of equations of the form: (1). y = m1 x + b1
(2). y = m2 x + b2
If m1 = m2 and b1 = b2, then the two lines are parallel and coinciding, and the system has an infinite number of solutions.
Unit 3 Solving a System of Equations by Substitution
To solve a system of equations of the first degree in two variables by the substitution method:
1. Isolate one variable as a function of the other in only one of the two equations in the system.
2. Substitute the expression obtained for this variable in the other equation in the system.
3. Solve this equation in one variable.
4. Substitute this value in the equation obtained in step 1.
5. Test the solution in each original equation.
6. State the solution.
Unit 4 Solving a System of Equations by Elimination through Addition
Equivalent equations of the first degree are represented graphically by the same line.
Equivalent equations can be obtained:
• By adding or subtracting the same amount from each of the two members of an equation according to the addition rule for equality.
• By multiplying or by dividing each of the two members of the equation by the same amount
(except 0), according to the multiplication rule for equality.
Two systems of equations are equivalent when the solutions of the first system are also the solutions of the second.
Two numbers are said to be opposite or additive inverses if they are formed by the same digits but have opposite signs.
To solve a system of equations of the first degree by elimination through addition:
1. Choose a variable to be eliminated.
2. Transform the equations in the system to equivalent equations, in such a way that the coefficients of the variable to be eliminated are inverses.
3. Add the equations in the system member to member.
4. Solve the equation obtained.
5. Substitute the value obtained in one or other of the equations.
6. Test the solution in each equation in the system.
7. State the solution.
Unit 5 Solving a System of Equations: Four Possible Methods
Graphing
Comparison
Substitution
Elimination through addition
The Rule of Transposition states that a term can be transferred from one side of an equation to the other side provided its sign is reversed.
In a proportion, the product of the means is equal to the product of the extremes. That is, if
[pic] then [pic] (cross multiplication)
Note that equations of the first degree in two variables are the equations of straight lines and when you solve them simultaneously, you are finding the value of x and the value of y that satisfy both equations. That is you are finding the coordinates of the point where the lines cross. If the lines are parallel or if both equations represent the same line then this point does not exist and there is no simultaneous solution.
Examples of the algebraic methods to solve systems of linear first-degree equations:
To solve a system of equations of the first degree in two variables graphically:
1. Transform each of the equations in the system to the form y = m x + b.
2. Determine at least three solution-pairs for each equation and complete a table of values.
3. Graph each equation in the same Cartesian plane.
4. Identify the point of intersection of the two lines, if there is one.
5. Determine the solution of the system of equations.
Example of the Method of Comparison:
Solve the following system of equations.
(1) 3y + 3x – 45 = 0
(2) 2y = – ([pic]) x + 24
Arrange the equations to get the variable y on the left-hand side and all other terms on the right-hand side.
(1) 3y = –3x + 45
(2) 2y = – ([pic]) x + 24
Now in both equation, isolate the variable y to get the equations in the form y = m x + b.
(1) y = –x + 15
(2) y = – ([pic]) x + 12
(–x + 15 = – ([pic]) x + 12 {comparing these two equal expressions and solving for the variable x}
(–x + ([pic]) x = 12 – 15 {transposing the x terms to the left hand side and the numerical terms to the right hand side}
((–1 +[pic]) x = – 3
(([pic]) x = – 3 {dividing both sides by[pic]}
(x = [pic] = 9
Substitute this value of x in equation (1)
(y = –9 + 15 = 6
Solution set is (x, y) = ( 9 , 6)
Put these values into equations (1) and (2) to verify the solution.
Example of the Method of Substitution:
Solve the following system of equations.
(1) 5x + y = 8
(2) y – 5x = 6
In equation (2) obtain the value of y by transposing –5x to the other side of the equation and changing its sign (Rule of Transposition).
(3) y = 5x + 6
Now substitute this value of y into equation (1) to get
(4) 5x + (5x + 6) = 8
(10x + 6 = 8
(10x = 8 ( 6 = 2
(x = [pic]
Now substitute this value of x in equation (2) to get
5 ([pic]) + y = 8
(1 + y = 8
(y = 8 ( 1 = 7 Solution set is (x, y) = ([pic], 7)
Put these values into equations (1) and (2) to verify the solution.
Example 1 of the Method of Elimination by Addition :
Solve the following system of equations.
(1) 6x + 4y = 30
(2) 3x + y = 12
Multiply equation (2) by (4 to get
(3) (12x ( 4y = ( 48 Now we can eliminate the y variable by adding
i.e. add equations (1) and (3) to get
( 6x = ( 18
(x = [pic] = 3
now substitute this value of x in equation (2) to get
3(3) + y = 12
(9 + y = 12
( y = 12 ( 9 = 3
Solution set is (x, y) = ( 3 , 3)
Put these values into equations (1) and (2) to verify the solution.
Example 2 of the Method of Elimination by Addition :
Solve the following system of equations.
(1) 2x – 3y = 7
(2) –x + 2 = 9(y + 1)
Expand the terms in equation (2) and bring the y term over to the left-hand side and the constants to the right hand side to get equation (3) then (4)
(3) –x + 2 = 9y + 9
(4) –x – 9y = –2 + 9 = 7
Now multiply along in equation (4) by 2 to get
(5) –2x – 18y = 14
Now add equations (1) and (5) to eliminate the x term and get
–21y = 21
(y = [pic]= –1
Substitute this value of y in equation (1) to get
2x – 3(–1) = 7
(2x + 3 = 7
(2x = 7 – 3 = 4
(x = [pic] = 2
Solution set is (x , y) = ( 2 , –1)
Put these values into equations (1) and (2) to verify the solution.
Unit 6 Solving Everyday Problems
Translate statements into algebraic terms.
To translate everyday situations into mathematical language :
1. Read the problem carefully and identify the pertinent information in the problem. That is carefully identify, then write down, all the information provided.
2. Identify the unknown variables. The algebraic variables x and y are normally used to represent the values required.
3. State the problem forming it as a system of algebraic equations.
4. Solve the problem by one of the methods studied i.e. graphically, by comparison,
by substitution, or by elimination through addition.
5. Verify the solution by substituting the obtained values for x and y in the each of the original system of equations and by checking whether the solution makes sense by comparing it with the problem.
Unit 7 Graphing a System of Inequalities
An Inequality is a sentence formed by two mathematical expressions (called members of the inequality) containing one or more variables which are related by an inequality symbol.
≥ is greater than or equal to
≤ is less than or equal to
> is greater than
< is less than
The solution of an inequality of the first degree in two variables is made up of ordered pairs (x , y ) which satisfy the inequality. Thus all ordered pairs (x , y ) that satisfy the inequality x + y ≥ 8 are solution pairs of this inequality. e.g. (6 , 3) , (4 , 7) , (4 , 4) …….
A line separates a Cartesian Plane into two half-planes. These half planes correspond to an inequality of the first degree in two variables. The line represents the boundary between these two regions and it corresponds to an equation of the first degree in two variables.
To represent an inequality of the first degree in two variables:
1. Graph the line of the equation associated with this inequality. This line represents the boundary of the required half-plane.
2. Choose any point located outside the line and note the ordered pair representing .this point.
3. Test whether the chosen ordered pair is a solution of the inequality.
4. Shade or cross-hatch the half plane corresponding to the inequality.
Closed half-planes are identified by inequalities which include the equal sign, i.e. by the symbols
≥ or ≤. The line belongs to the solution set if it is represented by a solid line.
Open half-planes are identified by inequalities which exclude the equal sign, i.e. by the symbols
> or ................
................
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