In this lesson, we learned how to solve for variables through substitution. The first step is to isolate the variable. Once you have determined its value, substitute it for the same variable in the second equation. By basic computation, you should be able to identify the value of each variable. If the solutiom is, for example, 2 = 8, then write "no solution." If the solution is, for example, 3 = 3, then write "all real solutions." The following is an example of how to perform substitution:
x + y = 4, x + y = -2
1) x = 4 - y
2) (4 - y) - y = -2
4 - 2y = -2
-2y = -6
y = 3
3) x + 3 = 4
x = 1
Section 7.2: Systems of Linear Equations in 2 Variables
In this lesson, we learned how to solve for variables through elimination. First, obtain coefficients that differ only in sign. Second, add equations to eliminate a variable. Then back substitute to solve for the second equation. Finally, check your solution! There are three possible types of solutions: intersecting lines (one solution), parallel lines (no solution), and the same line (infinitely many solutions). The following is an example of how to perform elimination:
5x + 3y = 9
2x - 4y = 14
Step 1:
4(5x + 3y = 9) --> 20x + 12y = 36
3(2x - 4y = 14) --> 6x - 12y = 42
Step 2:
20x + 12y = 36
6x - 12y = 42
+ -----------------
26x = 78
x = 3
Step 3:
2(3) - 4y = 14
6 - 4y = 14
-4y = 8
y = -2
Step 4:
2(3) - 4(-2) = 14
6 + 8 = 14
14 = 14
(3, -2)
Section 7.3: Multivariable Linear Systems
The following are methods to solve 3 equations with 3 unknown variables:
Using Back-Substitution
x - 2y + 3z = 9
y + 3z = 5
z = 2
1) Substitute 2 for z in the second equation.
y + 3(2) = 5
y = -1
2) Substitute -1 for y and 2 for z in the first equation.
x - 2(-1) + 3(2) = 9
x = 1
3) Write your solution as an ordered pair.
(1, -1, 2)
Using Elimination
x - 2y + 3z = 9
-x + 3y + 0z = -4
2x - 5y + 5z = 17
First, eliminate all the x's from the first column.
1) Add the first equation to the second equation, producing a new second equation.
x - 2y + 3z = 9
-x + 3y + 0z = -4
+ -----------------
y + 3z = 5
2) Add -2 times the first equation to the third equation, producing a new third equation.
-2(x - 2y + 3z = 9)
2x - 5y + 5z = 17
+ -----------------
-y - z = -1
Second, eliminate all the y's from the second column.
3) Add the second equation to the third equation, producing a new third equation.
x - 2y + 3z = 9
y + 3z = 5
2z = 4
Finally, solve for z in the third equation.
4) Multiply the third equation by 1/2, producing a new third equation.
x - 2y + 3z = 9
y + 3z = 5
z = 2
5) Write your solution as an ordered pair.
(1, -1, 2)
Nonsquare systems occur when there are 3 unknown variables and only 2 equations.
x - 2y + z = 2
2x - y - z = -3
1) Add -2 times the first equation to the second equation, producing a new second equation.
x - 2y + z = 2
3y - 3z = -3
2) Multiply the second equation by 1/3, producing a new second equation.
x -2y + z = 2
y - z = -1
3) Solving for y in terms of z, you get y = z - 1, and back-substitution into equation 1 yields
x - 2(z - 1) + z = 2
x - 2z + 2 + z = 2
x = z
4) Finally, by letting z = a, you have the solution written as an ordered pair
(a, a - 1, a)
Section 7.4: Partial Fractions
In this lesson, we learned how to perform partial fraction decomposition, which involves writing a rational expression as the sum of two or more rational expressions. For instance, (x+7) / (x^2-x-6) can be written as the sum of two fractions with linear denominators. That is, (x+7) / (x^2-x-6) = (2 / x-3) + (-1 / x+2). Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition of the left side. The following are steps to carry out partial fraction decomposition:
1. Multiply by LCD
2. Distribute
3. Collect like terms
4. Factor out the variable
5. Equate the coefficients
6. Solve the system of equations
7. Write the answer as a partial fraction
When the denominator of a partial fraction is, for example, (x+1)^2, it is called a repeated factor.
When the denominator of a partial fraction is, for example, (x^2+1), it is called a quadratic factor.
Here are a few examples:
Example 1: Distinct Linear Factors
5/(x^2+x-6)--> 5/(x+3)(x-2) = A/(x+3) + B/(x-2)
1. 5 = A(x-2) + B(x+3)
2. 5 = Ax - 2A + Bx + 3B
3. 5 = Ax + Bx - 2A + 3B
4. 5 = (A + B)x - 2A + 3B
5. 0 = A + B, 5 = -2A + 3B
6. 5 = 5B--> B = 1, A = -1
7. -1/(x+3) + 1/(x-2)
Example 2:Repeated Linear Factors
2x - 3/(x-1)^2
In this case, a factor (x-1) repeats, and thus the steps to solve it are as follows:
1. Multiply by LCD
2. Distribute
3. Collect like terms
4. Equate the coefficients
5. Solve the system of equations
6. Write the answer as a partial fraction
2x - 3/(x-1)^2 = A/(x-1) + B/(x-1)^2 Every exponent gets a factor!
1. 2x-3 = A(x-1) + B
2. 2x-3 = Ax - A + B
3. 2x-3 = Ax + (-A + B)
4. A = 2, -3 = -A + B
5. -3 + 2 = B = -1
6. 2/(x-1) - 1/(x-1)^2
Example 3: Distinct Linear Factors
4x^2 + 2x - 1 / x^2(x +1) = A/x + B/x^2 + C/x+1
1. 4x^2 + 2x - 1 = Ax(x+1) + B(x+1) + Cx^2
2. 4x^2 + 2x - 1 = Ax^2 + Ax + Bx + B + Cx^2
3. 4x^2 + 2x - 1 = Ax^2 + Cx^2 + Ax + Bx + B
4. 4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B
5. A + C = 4, A + B = 2, B = -1
6. A = 3, B = -1, C = 1
7. 3/x -1/x^2 + 1/x+1
Example 4: Distinct Linear and Quadratic Factors
x^2 - 1 / x(x^2 + 1)
In this case, the denominator contains an irreducible quadratic factor, so the set up is as follows:
x^2 - 1 / x(x^2 + 1) = A/x + Bx + C / x^2 + 1
1. x^2 - 1 = A(x^2 + 1) + (Bx + C)x
2. x^2 - 1 = Ax^2 + A + Bx^2 + Cx
3. x^2 - 1 = Ax^2 + Bx^2 + Cx + A
4. x^2 - 1 = (A + B)x^2 + Cx + A
5. A + B = 1, C = 0, A = -1
6. A = -1, B = 2, C = 0
7. -1/x + 2x/ x^2 + 1
When the exponent of the numerator is greater than or equal to the exponent of the denominator, use long division. Once you have found the quotient, take the remainder and put it over the original denominator, using it as your new partial fraction.
Section 7.5: Systems of Inequalities
In this lesson, we learned how to solve and graph systems of inequalities. The graph of the equation will normally separate the plane into two or more regions. In each region, one of the following must be true: a) All points in the region are solutions of the inequality. b) No point in the region is a solution of the inequality. Thus, you can determine whether the points in an entire region satisfy the inequality by simply testing one point in the region. The following are steps to graph an inequality in 2 variables:
1. Replace the inequality sign with an equal sign, and sketch the graph of the resulting equation.
(Use a dashed line for < or > and a solid line for <(=) or >(=)).
2. Test one point in each of the regions formed by the graph in Step 1. If the point satisfies the inequality, shade the entire region todenote that every point in the region satisfies the inequality.
3. A solution of a system of inequalities in x and y is a point (x , y) that satisfies each inequality.
4. For a system of inequalities, it is helpful to find the vertices of the solution region.
Here are some examples:
Example 1:
Sketch the graph of y >(=) x^2 - 1
The graph, as shown above, of the corresponding equation y = x^2 - 1 is a parabola. By testing the point (0,0) above the parabola and the point (0,-2) below the parabola, you can see that the points that satisfy the inequality are those lying above (or on) the parabola. This inequality is also nonlinear.
Example 2:
Sketch the graphs of x >(=) -2 and y <(=) 3.
a) The graph of the corresponding equation x = -2 is a vertical line. The points that satisfy the inequality are those lying to the right of this line. b) The graph of the corresponding equation y = 3 is a horizontal line. The points that satisfy the inequality are those lying below (or on) this line.
Example 3:
Sketch the region containing all points that satisfy the system.
x^2 - y <(=) 1
-x + y <(=) 1
As shown in the graph above, the points that satisfy the inequality x^2 - y <(=) 1 are the points lying above (or on) the parabola given by y = x^2 - 1. The points that satisfy the inequality -x + y <(=) are rhe points lying below (or on) the line given by y = x + 1. To find the points of the intersection of the parabola and line, solve the system of equations. Using the method of substitution, you will find the solutions to be (-1,0) and (2,3).
Lesson 7.6: Linear Programming
In this lesson, we learned how to perform linear programming. A 2D linear programming problem consists of a linear objective function and a system of linear inequalities called constraints. The objective function gives the quantity that is to be maximized or minimized, and the constraints determine the set of feasible solutions.
The following are steps to solving a linear programming problem:
1) Sketch the region corresponding to the system of constraints.
2) Find the vertices of the region.
3) Test the objective function, z = ax + by, at each of the vertices and select the values of the variables that optimize the objective function. For a bounded region, both a maximum and minimum will exist. For an unbound region, if an optimized solution exists, it will occur at a vertex.
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