GRAPHICAL SOLUTIONS
Often, we want to find a single ordered pair that is a solution to two different linearequations. One way to obtain such an ordered pair is by graphing the two equationson the same set of axes and determining the coordinates of the point where theyintersect.
Example 1
Graph the equations
x + y = 5
x - y = 1
on the same set of axes and determine the ordered pair that is a solution for eachequation.
Solution
Using the intercept method of graphing, we find that two ordered pairs that aresolutions 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 methodsfor exact solutions in later sections.
Linear equations considered together in this fashion are said to form a system ofequations. As in the above example, the solution of a system of linear equationscan be a single ordered pair. The components of this ordered pair satisfy each ofthe 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 intersect-that is, if the linesare parallel (see Figure 8.1a)-the equations are said to be inconsistent, and thereis no ordered pair that will satisfy both equations. If the graphs of the equations arethe same line (see Figure 8.1b), the equations are said to be dependent, and eachordered pair which satisfies one equation will satisfy both equations. Notice thatwhen a system is inconsistent, the slopes of the lines are the same but they-intercepts are different. When a system is dependent, the slopes and y-interceptsare the same.
In our work we will be primarily interested in systems that have one and only onesolution and that are said to be consistent and independent. The graph of such asystem is shown in the solution of Example 1.
SOLVING SYSTEMS BY ADDITION I
We can solve systems of equations algebraically. What is more, the solutions weobtain by algebraic methods are exact.
The system in the following example is the system we considered in Section 8.1on page 335.
Example 1
Solve
x + y = 5 (1)
x - y = 1 (2)
Solution
We can obtain an equation in one variable by adding Equations (1) and (2)
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 thecorresponding 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 canobtain an equation containing a single variable. The equation in one variable,together with either of the original equations, then forms an equivalent systemwhose solution is easily obtained.
In the above example, we were able to obtain an equation in one variable byadding Equations (1) and (2) because the terms +y and -y are the negatives of eachother. Sometimes, it is necessary to multiply each member of one of the equationsby -1 so that terms in the same variable will have opposite signs.
Example 2
Solve
2a + b = 4 (3)
a + b = 3(4)
Solution
We begin by multiplying each member of Equation (4) by - 1, to obtain
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, anequation 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, theordered pair is given in the form (a, b).
SOLVING SYSTEMS BY ADDITION II
As we saw in Section 8.2, solving a system of equations by addition depends onone of the variables in both equations having coefficients that are the negatives ofeach other. If this is not the case, we can find equivalent equations that do havevariables with such coefficients.
Example 1
Solve the system
-5x + 3y = -11
-7x - 2y = -3
Solution
If we multiply each member of Equation (1) by 2 and each member of Equation(2)by 3, we obtain the equivalent system
(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 theleft-hand member and the constant term is in the right-hand member. We will referto such arrangements as the standard form for systems. It is convenient to arrangesystems in standard form before proceeding with their solution. For example, if wewant to solve the system
3y =5x-11
-7x =2y-3
we would first write the system in standard form by adding -5x to each memberof 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 SUBSTITUTION
In Sections 8.2 and 8.3, we solved systems of first-degree 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 usingEquation (1). But we also could have used Equation (2) to write x explicitly in termsof 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 variablein terms of the other variable.
APPLICATIONS USING TWO VARIABLES
If two variables are related by a single first-degree equation, there are infinitelymany ordered pairs that are solutions of the equation. But if the two variables arerelated by two independent first-degree equations, there can be only one orderedpair that is a solution of both equations. Therefore, to solve problems using twovariables, we must represent two independent relationships using two equations.We can often solve problems more easily by using a system of equations than byusing a single equation involving one variable. We will follow the six steps outlinedon 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 thesmaller number. Find the numbers.
Solution
Steps 1-2
We represent what we want to find as two word phrases. Then, werepresent the word phrases in terms of two variables.
Smaller number: x
Larger number: y
Step 3 A sketch is not applicable.
Step 4 Now we must write two equations representing the conditions stated.
The sum of two numbers is 26.
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 bythe 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
Two equations considered together form a system of equations. The solution isgenerally a single ordered pair. If the graphs of the equations are parallel lines, theequations are said to be inconsistent; if the graphs are the same line, the equationsare said to be dependent.
We can solve a system of equations by the addition method if we first write thesystem in standard form, in which the terms involving the variables are in theleft-hand member and the constant term is in the right-hand member.
We can solve a system of equations by the substitution method if one variable inat least one equation in the system is first expressed explicitly in terms of the othervariable.
We can solve word problems using two variables by representing two independentrelationships by two equations.
