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SYSTEMS OF LINEAR EQUATIONSGRAPHICAL SOLUTIONSOften, we want to find a single ordered pair that is a solution to two different linear equations. One way to obtain such an ordered pair is by graphing the two equations on the same set of axes and determining the coordinates of the point where they intersect. Example 1 Graph the equations
on the same set of axes and determine the ordered pair that is a solution for each equation. Solution Using the intercept method of graphing, we find that two ordered pairs that are solutions of x + y = 5 are (0, 5) and (5, 0) And two ordered pairs that are solutions of x  y = 1 are (0,1) and (1,0) The graphs of the equations are shown. The point of intersection is (3, 2). Thus, (3, 2) should satisfy each equation. In fact, 3 + 2 = 5 and 3  2 = 1 In general, graphical solutions are only approximate. We will develop methods for exact solutions in later sections. Linear equations considered together in this fashion are said to form a system of equations. As in the above example, the solution of a system of linear equations can be a single ordered pair. The components of this ordered pair satisfy each of the two equations. Some systems have no solutions, while others have an infinite number of solu tions. If the graphs of the equations in a system do not intersectthat is, if the lines are parallel (see Figure 8.1a)the equations are said to be inconsistent, and there is no ordered pair that will satisfy both equations. If the graphs of the equations are the same line (see Figure 8.1b), the equations are said to be dependent, and each ordered pair which satisfies one equation will satisfy both equations. Notice that when a system is inconsistent, the slopes of the lines are the same but the yintercepts are different. When a system is dependent, the slopes and yintercepts are the same. In our work we will be primarily interested in systems that have one and only one solution and that are said to be consistent and independent. The graph of such a system is shown in the solution of Example 1. SOLVING SYSTEMS BY ADDITION IWe can solve systems of equations algebraically. What is more, the solutions we obtain by algebraic methods are exact. The system in the following example is the system we considered in Section 8.1 on page 335. Example 1 Solve
Solution Solving the resulting equation for x yields 2x = 6, x = 3 We can now substitute 3 for x in either Equation (1) or Equation (2) to obtain the corresponding value of y. In this case, we have selected Equation (1) and obtain (3) + y = 5 y = 2 Thus, the solution is x = 3, y = 2; or (3, 2). Notice that we are simply applying the addition property of equality so we can obtain an equation containing a single variable. The equation in one variable, together with either of the original equations, then forms an equivalent system whose solution is easily obtained. In the above example, we were able to obtain an equation in one variable by adding Equations (1) and (2) because the terms +y and y are the negatives of each other. Sometimes, it is necessary to multiply each member of one of the equations by 1 so that terms in the same variable will have opposite signs. Example 2 Solve 2a + b = 4 (3) a + b = 3 (4) Solution
2a + b = 4 (3) a  b =  3 (4') where +b and b are negatives of each other. The symbol ', called "prime," indicates an equivalent equation; that is, an equation that has the same solutions as the original equation. Thus, Equation (4') is equivalent to Equation (4). Now adding Equations (3) and (4'), we get Substituting 1 for a in Equation (3) or Equation (4) [say, Equation (4)], we obtain 1 + b = 3 b = 2 and our solution is a = 1, b = 2 or (1, 2). When the variables are a and b, the ordered pair is given in the form (a, b). SOLVING SYSTEMS BY ADDITION IIAs we saw in Section 8.2, solving a system of equations by addition depends on one of the variables in both equations having coefficients that are the negatives of each other. If this is not the case, we can find equivalent equations that do have variables with such coefficients. Example 1 Solve the system 5x + 3y = 11 7x  2y = 3 Solution
(2) (5x) + (2)(3y) = (2)(ll) (3) (7x)  (3)(2y) = (3)(3) or 10x + 6y = 22 (1') 21x  6y = 9 (2') Now, adding Equations (1') and (2'), we get 31x = 31 x = 1 Substituting 1 for x in Equation (1) yields 5(1) + 3y = 11 3y = 6 y = 2 The solution is x = 1, y = 2 or (1, 2). Note that in Equations (1) and (2), the terms involving variables are in the lefthand member and the constant term is in the righthand member. We will refer to such arrangements as the standard form for systems. It is convenient to arrange systems in standard form before proceeding with their solution. For example, if we want to solve the system 3y = 5x  11 7x = 2y  3 we would first write the system in standard form by adding 5x to each member of Equation (3) and by adding 2y to each member of Equation (4). Thus, we get 5x + 3y = 11 lx  2y = 3 and we can now proceed as shown above. SOLVING SYSTEMS BY SUBSTITUTIONIn Sections 8.2 and 8.3, we solved systems of firstdegree equations in two vari ables by the addition method. Another method, called the substitution method, can also be used to solve such systems. Example 1 Solve the system 2x + y = 1 (1) x + 2y = 17 (2) Solution Solving Equation (1) for y in terms of x, we obtain y = 2x + 1 (1') We can now substitute 2x + 1 for y in Equation (2) to obtain x + 2(2x + 1) = 17 x + 4x + 2 = 17 5x = 15 x = 3 (continued) Substituting 3 for x in Equation (1'), we have y = 2(3) + 1 = 7 Thus, the solution of the system is a: x = 3, y = 7; or (3, 7). In the above example, it was easy to express y explicitly in terms of x using Equation (1). But we also could have used Equation (2) to write x explicitly in terms of y x = 2y + 17 (2') Now substituting  2y + 17 for x in Equation (1), we get Substituting 7 for y in Equation (2'), we have x = 2(7) + 17 = 3 The solution of the system is again (3, 7). Note that the substitution method is useful if we can easily express one variable in terms of the other variable. APPLICATIONS USING TWO VARIABLESIf two variables are related by a single firstdegree equation, there are infinitely many ordered pairs that are solutions of the equation. But if the two variables are related by two independent firstdegree equations, there can be only one ordered pair that is a solution of both equations. Therefore, to solve problems using two variables, we must represent two independent relationships using two equations. We can often solve problems more easily by using a system of equations than by using a single equation involving one variable. We will follow the six steps outlined on page 115, with minor modifications as shown in the next example. Example 1 The sum of two numbers is 26. The larger number is 2 more than three times the smaller number. Find the numbers. Solution
Steps 12 Step 3 A sketch is not applicable. Step 4 Now we must write two equations representing the conditions stated.
Step 5 To find the numbers, we solve the system x + y = 26 (1) y = 2 + 3x (2) Since Equation (2) shows y explicitly in terms of x, we will solve the system by the substitution method. Substituting 2 + 3x for y in Equation (1), we get x + (2 + 3x) = 26 4x = 24 x = 6 Substituting 6 for x in Equation (2), we get y = 2 + 3(6) = 20 Step 6 The smaller number is 6 and the larger number is 20. CHAPTER SUMMARY
