Definitions:
System of Linear Equations: 2 or more linear equations
System of Linear Inequalities: 2 or more linear inequalities
Solution of the System of Linear Equations: any ordered pair in a system that makes all equations true.
Solution of the System of Linear Inequalities: any ordered pair in a system that makes all inequalities true.
Key Concepts:
One Solution: Different Slopes
No Solution: Lines are parallel, same slopes but different y-intercepts
Infinitely Many Solutions: Lines are the same, same slope and same y-intercepts
Solving System of Equations by Graphing
In order to solve a system of equations by graphing Slope-Intercept Form (y = mx + b):
o Graph each equation using the y-intercept (b) and the slope (m).
Standard Form (Ax + By = C) o Find the slope (-A/B)
o Find the x-intercepts & y-intercepts
o Graph using the slope, x-intercepts & y-intercepts
In order to ensure that both lines are long and will connect, make sure to plot the slope of each line several times in both directions.
The point where 2 lines intersect is the solution the system.
Check to see if the solution makes both equations true. Examples: Solve System of Equations by Graphing
Solve by graphing. Check your solution. 1. y = x + 4
y = 4x + 1 Slope Intercept Form:
Step 1: Find slope & y-intercepts b1 = (0, 4) m1 = (1↑ 1→)
b2 = (0, 1) m2 = (4↑ 1→) Step 2: Determine # of solutions
Since slopes and y-intercepts are different, there is 1 solution.
Step 3: Graph equations & look for point of intersection
Step 4: Check solution (1, 5) y = x + 4 y = 4x + 1 5 = (1)+4 5 = 4(1) + 1 5= 5 5 = 4 + 1 5 = 5 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y
2. –x + y = 5 4x + y = 0 Standard Form:
Step 1: Find slope, x-intercepts & y-intercepts b1 = (0, 5) m1 = 1 1 (1↑ 1→) b2 = (0, 0) m2 = 4 1 (4↓1→) Step 2: Determine # of solutions
Since slopes and y-intercepts are different, there is 1 solution.
Step 3: Graph equations & look for point of intersection
Step 4:
Check solution (-1, 4)
-x + y = 5 4x + y = 0 -(-1)+ 4 = 5 4(-1) + 4 = 0 1 + 4 = 5 -4 + 4 = 0 Practice: Solve System of Equations by Graphing
Solve by graphing. Check your solution. 3. y = 2x + 1 y = 3x – 1 Solution: 4. 2x + 4y = 8 3x + 4y = 12 Solution:
Example: Word Problem
5. Suppose you have $20 in your bank account. You start saving $5 each week. Your friend has $5 in his account and is saving $10 each week. Assume that neither you nor your friend makes any withdrawals.
a. After how many weeks will you and your friend have the same amount of money in your accounts?
b. How much money will each of you have?
1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y
Solving System of Inequalities by Graphing
In order to solve a system of inequalities by graphing Repeat the same steps for graphing equations
Use the inequality symbol to draw either dotted or solid lines and shade the appropriate area.
The area that is shaded by both inequalities is the solution to the system of inequalities.
Check to see if the solution makes both inequalities true. Example: Solve System of Inequalities by Graphing
Solve by graphing. Check your solution. 6. y > 2x – 3
2x + 1y < 4
Step 1: Find relevant information to graph Ineq 1: Slope Intercept Form:
Find slope & y-intercepts b1 = (0, -3) m1 =
2
1 (2↑ 1→) Ineq 2: Standard Form:
Find slope, x-intercepts & y-intercepts b2 = (0,4) m2 =
-2
1 (2↓ 1→), x-int = (2,0)
Step 2: Graph inequalities & look for overlapping region
Step 3: Find a solution & check it.
Practice: Solve System of Inequalities by Graphing 7. y < 2x + 4 2x - y ≤ 4 8. y > 1 4x y ≤ -x + 4 9. y ≥ -x + 5 y ≤ 3x - 4 10. y ≥ -x + 1 y ≤ -1 3x + 5 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y 1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 4 5 –1 –2 –3 –4 –5 y
Evaluating Expressions
When asked to evaluate an expression, you simply replace the variable with the value and follow your order of operations (PEMDAS)
Examples: Evaluating Expressions
Evaluate the following expressions for the given values. 11. Y = -x + 4 for x = 5
Step 1: replace x with 5 Y = -(5) + 4
Step 2: Follow PEMDAS and simplify Y = -5 + 4 = -1
12. Y = x – 6 for x = 5
13. Y = -x + 4 for y = 1 14. Y = x - 6 for y = 1
15. 4x + 5y = 20 for x = 2 16. 4x + 5y = 20 for y = 5
Solving for a variable
When asked to solve an equation for an indicated variable, Follow same steps as solving an equation.
Note: result will typically be an algebraic expression
Examples: Solving for indicated variables
Solve each equation for the indicated variables 17. Y = -x + 4 solve for x 18. d = a x 2 + b solve for x
19. 3y + 6x = 12 solve for y 20. ax + bx -15 = 0 solve for x
Substitution Method
Replace 1 variable with an equivalent expression containing the other variable
Solve for other variable using process for solving a 1 variable equation
Substitute value of solved variable to find the value of the other variable.
Check solution
Example: Solve by Substitution
23. y = -4x + 8y = x + 7
Step 1: Write equation containing 1 variable & solve. y = -4x + 8
x + 7 = -4x + 8 Substitute x + 7 for y +4x +4x Add 4x to both sides
5x + 7 = 8
-7 -7 Subtract 7 from both sides 5x = 1
5 5 Divide both sides by 5 x = 0.2
Step 2: Solve for other variable in either equation. Step 3: Check solution (0.2, 7.2)
y = x + 7 7.2 = -4(0.2) + 8 0.2 + 7 = 7.2
y = 0.2 + 7 (Substitute 0.2 for x) 7.2 = -.8 + 8 = 7.2 7.2 = 7.2 y = 7.2
24. 3y + 2x = 4 -6x + y = -7
Step 1: Solve 1 variable in terms of the other variable. -6x + y = -7
+6x +6x Add 6x to both sides y = 6x -7
Step 2: Write equation containing 1 variable & solve. 3(6x – 7) + 2x = 4 Substitute 6x – 7 for y in 3y + 2x = 4. 18x – 21 + 2x = 4 Use distributive property.
20x – 21 = 4 Combine like terms +21 +21 Add 21 to both sides
20x = 25
20 20 Divide both sides by 20
x = 1.25
Step 3: Solve for other variable in either equation. Step 4: Check solution (1.25, 0.5)
-6x + 7 = 7 3y + 2x = 4 -6x + y = -7
-6(1.25) + y = -7 Substitute 1.25 for x 3(0.5) + 2(1.25) = 4 -6(1.25) + 0.5 = -7 -7.5 + y = -7 1.5 + 2.5 = 4 -7.5 + 0.5 = -7 +7.5 +7.5 Add 7.5 to both sides. 4 = 4 -7 = -7
Practice: Solve by Substitution
25. y = 2x 7x – y = 15 26. y = 2x + 2 y = -3x + 4 27. –2x + y = -1 4x + 2y = 12 28. y = 2x 6x – y = 8 29. y = 3x + 1 x = 3y + 1 30. x – 3y = 14 x – 2 = 0Example/Practice: Real-World Problems
31. School committee is planning after-school trip by 193 people to competition at another school. There are 8 drivers available and 2 types of vehicles (school buses & minivans). School buses seat 51 people & minivans seat 8 people each. How many buses & minivans will be needed?
Step 1: Identify variables.
Let b = school buses and m = minivans Step 2: Write equations.
Drivers b + m = 8 People 51b + 8m = 193 Step 3: Solve algebraically.
Step 4: Check solution
32. A shop sells gift wrap for $4 per package and greeting cards for $10 per package. If the shop sells 205 packages in all and receives a total of $1084, how many packages of gift wrap and greeting cards were sold?
Elimination Method
Add or subtract equations to eliminate a variable.
o If not possible, then multiply 1 or both equations by number(s) that will allow you to add/subtract the equations.
Solve for the 2nd variable.
Plug in value of the 2nd variable into one of the equations to solve for the 1st variable.
Check solution
Example: Solve by Elimination (Add/Subtract Equations)
33. Solve by elimination:5x – 6y = -32 3x + 6y = 48
Step 1: Eliminate y because the sum of the coefficients of y = 0
5x – 6y = -32 + 3x + 6y = 48
8x + 0 = 16 (addition property of equality) x = 2 (Solve for x)
Step 2: Solve for eliminated variable using any of 2 equations.
3x + 6y = 48
3(2) + 6y = 48 (Substitute 2 for x)
6 + 6y = 48 (Simplify and then solve for y) y = 7
Step 3: Check Solution (2, 7)
5(2) – 6(7) = -32 3(2) + 6(7) = 48
10 – 42 = -32 6 + 42 = 48 -32 = -32 48 = 48
Practice: Solve by Elimination (Add/Subtract Equations)
34. 6x – 3y = 3 -6x + 5y = 3 35. 2x + 3y = 11 -2x + 9y = 1 36. 2x + 5y = 17 6x – 5y = -9
Example: Multiplying 1 Equation:
37. Solve by elimination: 2x + 5y = -22 10x + 3y = 22Step 1: Eliminate 1 variable
Start with given sys To prepare for eliminating x, multiply 1st eq by 5 Subtract equations to eliminate x. 2x + 5y = -22 5(2x + 5y = -22) 10x + 25y = -110
10x + 3y = 22 10x + 3y = 22 10x + 3y = 22
0 + 22y = -132 Step 2: Solve for y
22y = -132 y = -6
Step 3: Solve for eliminated variable using either equation: 2x + 5y = -22
2x + 5(-6) = -22 (Substitute –6 for y) 2x – 30 = -22
2x = 8
x = 4 (Solve for x) Step 4: Check solution (4, -6)
2(4) + 5(-6) = -22 10(4) + 3(-6) = 22
8 + -30 = -22 40 + -18 = 22
Practice: Multiplying 1 Equation:
38. –2x + 15y = -32 7x – 5y = 17 39. 3x + 6y = -6 -5x – 2y = -14 40. 16x + 8y = 0 8x + y = 24Example/Practice: Real-World Problems
41. EAHS sells a total of 292 tickets for a basketball game. An adult ticket is $3. A student ticket costs $1. $470 is collected in ticket sales. Write and solve a system to find the number of each type of ticket sold.
Step 1: Define the variables
Let a = # of adult tickets s = # of student tickets Step 2: Write 2 equations (1 for # of tickets & 1 for value of tickets)
a + s = 292 (total # of tickets) 3a + 1s = 470 (total amount of sales)
Step 3: Solve algebraically.
Step 4: Check solution
42. Tickets for a softball game at EAHS cost $5 for adults and $1 for students. The attendance that day was 1139, and $3067 was collected. Write and solve system of equations to find the number adults and number of students that attended the game.
Real-World Applications
When dealing with word problems, there will typically be 2 types of equations written
When given a total # and a value you will get
Quantity Equation (i.e. the total number of items you have)
Quality Equation (i.e. what is each item worth)
Depending on the 2 equations, the system can be solved in any of the following methods. Pick which one makes sense for the given problem
Solve by Graphing Solve by Substitution Solve by Elimination
Example:
43. You have 10 coins, all dimes and nickels. The value of the coins is $1.75. How many dimes and quarters do you have?
Let d = dimes and q = quarters Equation 1: Quantity Equation
d + q = 10
Equation 2: Quality Equation
Dimes are worth .10 and quarters are worth .25 .10d + .25q = 1.75
Looking at the 2 equations, I can easily use substitution.
Step 1: re-write equation 1 Step 2: replace q with 10-d in equation 2
Q = 10 – d .25(10 – d) + .10d = 1.75
Step 3: Solve for d Step 4: Solve for q
2.5 - .25d +.10d = 1.75 q + d = 10
-.15d = -.75 q + 5 = 10
d = 5 q = 5
Step 5: Check solution (5,5)
5 + 5 = 10 .25(5) + .10(5) = 1.75
10 = 10 1.25 + .50 = 1.75
1.75 = 1.75
44. An amusement park charges admission plus a fee for each ride. Admission plus 2 rides costs $10. Admission plus 5 rides cost $16. What is the charge for admission? For each ride?
Let a = admission, r = rides
Equation 1: a + 2r = 10 Equation 2: a + 5r = 16
Looking at the 2 equations, I can easily use elimination a + 2r = 10
- a + 5r = 16 -3r = -6 r = 2
Now solve for a: a + 2(2) = 10 => a + 4 = 10 => a = 6 Check solution (6,2)
Practice:
45. You have a total of 25 coins, all nickels and quarters. The total value is $2.85. Write and solve a system of equations to find the number of nickels (n) and number of quarters (q) that you have.
46. A local pizzeria sells a small pizza with 1 topping for $6.00 and a small pizza with 3 toppings for $8.00. What is the charge for a plain small pizza? What is the charge for each topping?
47. A new car dealerships sells cars and trucks in a ratio of 7 to 5. Last month the dealership sold 84 cars and trucks. How many cars and how many trucks were sold?
48. Suppose you bought supplies for a party. Three rolls of streamers and 15 party hats cost $30. Later, you bought 2 rolls of streamers and 4 party hats for $11. How much did each roll of streamers cost? How much did each party hat cost?
