1. Chapter 5 Review

2. Section 5.1 “Solve Linear Systems by Graphing”
Solving a Linear System by Graphing
(1) Graph both equations in the same plane.
(2) Estimate the coordinates of the point where the two lines intersect.
(3) Check the coordinate by substituting into EACH equation of the linear system, to see if the point is a solution for both equations.

3. Section 5.2 “Solve Linear Systems by Substitution”
Solving a Linear System by Substitution
(1) Solve one of the equations for one of its variables. (When possible, solve for a variable that has a coefficient of 1 or -1).
(2) Substitute the expression from step 1 into the other equation and solve for the other variable.
(3) Substitute the value from step 2 into the revised equation from step 1 and solve.

4. “Solve Linear Systems by Substituting”
y = 3x + 2
Equation 1
x + 2y = 11
Equation 2
x + 2(3x + 2) = 11
x + 2y = 11
Substitute
x + 6x + 4 = 11
7x + 4 = 11
x = 1
y = 3x + 2
Equation 1
Substitute value for
x into the original equation
y = 3(1) + 2
y = 5
(5) = 3(1) + 2
5 = 5
The solution is the point (1,5).
Substitute (1,5) into both equations to check.
(1) + 2(5) = 11
11 = 11

5. Section 5.3 “Solve Linear Systems by Elimination”
adding or subtracting equations to obtain a new equation in one variable.
Solving Linear Systems Using Elimination
(1) Add or Subtract the equations to eliminate one variable.
(2) Solve the resulting equation for the other variable.
(3) Substitute in either original equation to find the value of the eliminated variable.

6. + “Solve Linear Systems by Elimination Multiplying First!!”
Eliminated
x (2)
4x + 5y = 35
8x + 10y = 70
Equation 1 +
x (-5)
15x - 10y = 45
-3x + 2y = -9
Equation 2
23x = 115
x = 5
4x + 5y = 35
Equation 1
Substitute value for
x into either of the original equations
4(5) + 5y = 35
20 + 5y = 35
y = 3
4(5) + 5(3) = 35
35 = 35
The solution is the point (5,3).
Substitute (5,3) into both equations to check.
-3(5) + 2(3) = -9
-9 = -9

7. Section 5.4 “Solve Special Types of Linear Systems”
consists of two or more linear equations in the same variables.
Types of solutions:
(1) a single point of intersection – intersecting lines
(2) no solution – parallel lines
(3) infinitely many solutions – when two equations represent the same line

8. “Solve Linear Systems with No Solution”
Eliminated
Eliminated
3x + 2y = 10
Equation 1 _ +
-3x + (-2y) = -2
3x + 2y = 2
Equation 2
This is a false statement,
therefore the system has no
solution.
0 = 8
“Inconsistent
System”
No Solution
By looking at the graph, the lines
are PARALLEL and therefore will
never intersect.

9. “Solve Linear Systems with Infinitely Many Solutions”
Equation 1
x – 2y = -4
Equation 2
y = ½x + 2
Use ‘Substitution’ because we know what y is equals.
Equation 1
x – 2y = -4
x – 2(½x + 2) = -4
x – x – 4 = -4
This is a true statement,
therefore the system has infinitely many solutions.
-4 = -4
“Consistent
Dependent
System”
Infinitely Many Solutions
By looking at the graph, the lines are the SAME and therefore intersect at every point, INFINITELY!

10. Section 5.7 “Solve Systems of Linear Inequalities”
SYSTEM OF INEQUALITIES-
consists of two or more linear inequalities in the same variables.
x – y > 7
Inequality 1
2x + y < 8
Inequality 2
A solution to a system of inequalities is an ordered pair (a point) that is a solution to both linear inequalities.

11. Graph a System of Inequalities
y < x – 4
Inequality 1
y ≥ -x + 3
Inequality 2
Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darker shade of blue.
(5,0)
0 < 5 – 4 ?
0 ≥ ?
0 < 1
0 ≥ -2

12. Graph a System of THREE Inequalities
Inequality 1:
Graph all three inequalities in the same coordinate plane. The graph of the system is the triangular region, which is shown as the darker shade of blue.
x > -2
Inequality 2:
x + 2y ≤ 4
Inequality 3:
x + 2y ≤ 4 ?
y ≥ -1 ?
x > -2 ?
Check
(0,0)
0 ≥ -1
0 > -2
0 + 0 ≤ 4