Systems of Equations and Inequalities 9

Systems of Equations 9.1

Systems of Equations Here, we study how to solve systems of two equations in two unknowns. We learn three different methods of solving such systems: By substitution By elimination Graphically

Systems of Equations and their Solutions

A system of equations is a set of equations that involve the same variables. A solution of a system is an assignment of values for the variables that makes each equation in the system true. To solve a system means to find all solutions of the system.

Systems of Equations and their Solutions Here is an example of a system of two equations in two variables:

Systems of Equations and their Solutions We can check that x = 3 and y = 1 is a solution of this system. The solution can also be written as the ordered pair (3, 1). Equation 1Equation 2 2x – y = 5x + 4y = 7 2(3) – 1 = (1) = 7

Systems of Equations and their Solutions Note that the graphs of Equations 1 and 2 are lines. As the solution (3, 1) satisfies each equation, the point (3, 1) lies on each line. So, it is the point of intersection of the two lines.

Substitution Method

In the substitution method, we start with one equation in the system and solve for one variable in terms of the other variable. 1.Solve for one variable. 2.Substitute. 3.Back-substitute.

Step 1 Solve for one variable. Choose one equation and solve for one variable in terms of the other variable.

Step 2 Substitute. Substitute the expression you found in step 1 into the other equation to get an equation in one variable. Then, solve for that variable.

Step 3 Back-substitute. Substitute the value you found in step 2 back into the expression found in step 1 to solve for the remaining variable.

E.g. 2—Substitution Method Find all solutions of the system.

E.g. 2—Substitution Method Find all solutions of the system. We start by solving for y in the second equation. y = 3x – 10

E.g. 2—Substitution Method Next, we substitute for y in the first equation and solve for x: x 2 + (3x – 10) 2 = 100 x 2 + (9x 2 – 60x + 100) = x 2 – 60x = 0 10x(x – 6) = 0 x = 0 or x = 6

E.g. 2—Substitution Method Now, we back-substitute these values of x into the equation y = 3x – 10. For x = 0: y = 3(0) – 10 = –10 For x = 6: y = 3(6) – 10 = 8 So, we have two solutions: (0, –10) and (6, 8).

E.g. 2—Substitution Method The graph of the first equation is a circle. That of the second equation is a line. The graphs intersect at the two points (0, –10) and (6, 8).

Elimination Method

To solve a system using the elimination method, we try to combine the equations using sums or differences so as to eliminate one of the variables. 1.Adjust the coefficients. 2.Add the equations. 3.Back-substitute.

Step 1 Adjust the coefficients. Multiply one or more of the equations by appropriate numbers so that the coefficient of one variable in one equation is the negative of its coefficient in the other equation.

Step 2 Add the equations. Add the two equations to eliminate one variable. Then, solve for the remaining variable.

Step 3 Back-substitute. Substitute the value you found in step 2 back into one of the original equations. Then, solve for the remaining variable.

E.g. 4—Elimination Method Find all solutions of the system.

E.g. 4—Elimination Method Find all solutions of the system. We choose to eliminate the x-term. So, we multiply the first equation by 5 and the second equation by –3. Then, we add the two equations and solve for y.

E.g. 4—Elimination Method Now, we back-substitute y = –11 into one of the original equations (say, 3x 2 + 2y = 26) and solve for x.

E.g. 4—Elimination Method 3x 2 + 2(–11) = 26 3x 2 = 48 x 2 = 16 x = –4 or x = 4 So, we have two solutions: (–4, –11) and (4, –11)

E.g. 4—Elimination Method The graphs of both equations are parabolas. The graphs intersect at the two points (–4, –11) and (4, –11).

Graphical Method

In the graphical method, we use a graphing device to solve the system of equations. Note that, with many graphing devices, any equation must first be expressed in terms of one or more functions of the form y = f(x) before we can use the calculator to graph it.

Graphical Method Not all equations can be readily expressed in this way. So, not all systems can be solved by this method.

Graphical Method We proceed as follows. 1.Graph each equation. 2.Find the intersection points.

Step 1 Graph each equation. Express each equation in a form suitable for the graphing calculator by solving for y as a function of x. Graph the equations on the same screen.

Step 2 Find the intersection points. The solutions are the x- and y-coordinates of the points of intersection.

Graphical Method It may be more convenient to solve for x in terms of y in the equations. In that case, in step 1, graph x as a function of y instead.

E.g. 5—Graphical Method Find all solutions of the system.

E.g. 5—Graphical Method Find all solutions of the system. Solving for y in terms of x, we get the equivalent system

E.g. 5—Graphical Method The figure shows that the graphs of these equations intersect at two points. Zooming in, we see that the solutions are (–1, –1) and (3, 7).

E.g. 6—Solving a System of Equations Graphically Find all solutions of the system, correct to one decimal place. The graph of the first equation is a circle and the second a parabola. To graph the circle on a graphing calculator, we must first solve for y in terms of x (Section 2.3).

E.g. 6—Solving a System of Equations Graphically To graph the circle, we must graph both functions:

E.g. 6—Solving a System of Equations Graphically In the figure, the graph of the circle is shown in red and the parabola in blue.

E.g. 6—Solving a System of Equations Graphically The graphs intersect in quadrants I and II. Zooming in, or using the Intersect command, we see that the intersection points are: (–0.559, 3.419) and (2.847, 1.974)

E.g. 6—Solving a System of Equations Graphically There also appears to be an intersection point in quadrant IV.

E.g. 6—Solving a System of Equations Graphically However, when we zoom in, we see that the curves come close to each other but don’t intersect. Thus, the system has two solutions. Correct to the nearest tenth, they are: (–0.6, 3.4) (2.8, 2.0)