Solving Systems of Equations in Two Variables Section 2-1

Before finishing this section you should be able to: Solve systems of equations graphically Solve systems of equations algebraically

A system of equations is a set of two or more equations. Solving a system of equations means finding the that makes both equations true. This is called the solution to the system of equations. If a system of equations were graphed, the solution would be the point where the lines intersect. ordered pair

Solve the system of equations by graphing. 2x + 5y = 9 x - y = 1 First rewrite each equation of the system in slope-intercept form by solving for y. 2x + 5y = 9  y = x - y = 1  y = x - 1 Since the two lines have different slopes, the graphs of the equations are intersecting lines. The solution to the system is (2, 1)

A consistent system of equations has at least one solution. If there is exactly one solution, the system is independent. The previous example consistent (at least one solution) and independent (exactly one solution). Recall, the graphs of two equations may be the same line. In this case, there are infinitely many solutions and the system is dependent (also consistent because there is at least one solution). Lines may not always intersect. They do not intersect if the lines are parallel. Therefore, there is no solution and the system is inconsistent. The chart on the top of pg. 68 in your textbook summarizes the characteristics of these types of systems. Classifications of systems of equations:

A system of equations may be solved algebraically. Two common methods are the elimination method and the substitution method. The elimination method is when the equations are stacked and added together. One or both equations may have to be manipulated by multiplying the entire equation by a constant. The goal when adding the equations is that one of the variables is eliminated.

Solving by using Elimination Use the elimination method to solve the system of equations. 3x - 2y= 18 4x + 3y= -10 One way to solve this system is to multiply both sides of the first equation by 3, multiply both sides of the second equation by 2, and add the two equations to eliminate y. Then solve the resulting equation. Now substitute 2 for x in either of the original equations. The solution is (2, -6). Check it by substituting into 4x + 3y = -10. If the coordinates make both equations true, then the solution is correct.

In the SUBSTITUTION METHOD, first we must solve one of the equations for either x or y. (Hint: it will be easier to solve for the variable that has a coefficient of 1) Next, substitute what x equals into the other equation. Solve this equation for y. Substitute the solution for y into either of the two equations and solve for x.

Use the substitution method to solve the system of equations. 4x - 3y=11 x + y=8 Solving by using Substitution You can solve the second equation for either y or x. If you solve for x, the result is x = 8 - y. Then substitute 8 - y for x in the first equation. Now substitute 3 for y in either of the original equations, and solve for x. The solution is (5, 3)

Real-World Example CONSUMER CHOICES Jeremy is considering two different cell phone plans. The first plan has a $25 monthly fee plus $0.25 per minute used. The second plan offers a $10 monthly fee with a $0.40 charge per minute used. a.What is the break-even point in the two cell phone plans that Jeremy is considering? b.If Jeremy expects to use the phone for no more than 75 minutes each month, which plan should he choose?

a.First, write an equation to represent the amount he will pay with each plan. Let C represent the total monthly cost and m represent the number of minutes used. Plan 1 ($25 monthly charge plus $0.25 per minute):C = 0.25m + 25 Plan 2 ($10 monthly charge plus $0.40 per minute): C = 0.40m + 10 Now, solve the system of equations. Since both equations contain C, we can substitute the value of C from one equation into the other.

The break-even point occurs at a monthly usage of 100 minutes. b.The graph of the equations shows that for monthly usage under 100 minutes, Plan 2 is less expensive. So, Jeremy should probably choose Plan 2.

Helpful Websites Solving systems by graphing: Solving systems by algebra: V14-105_SAR1864,00.shtmlhttp:// V14-105_SAR1864,00.shtml s.clayton.edu/garrison/Math%200099/Solving_Systems_of_E quations.htmhttp://a- s.clayton.edu/garrison/Math%200099/Solving_Systems_of_E quations.htm 2-1 Self-Check Quiz: bin/msgQuiz.php4?isbn= &chapter=2&lesson=1&quizType=1&headerFile=4&state= bin/msgQuiz.php4?isbn= &chapter=2&lesson=1&quizType=1&headerFile=4&state