SYSTEMS OF LINEAR EQUATIONS
GRAPHICAL SOLUTIONS
Much, we want to find a single ordered pair that is a solution to 2 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
x + y = 5
x - y = 1
on the same set of axes and settle the ordered pair that is a solution for to each one equation.
Solution
Exploitation the intercept method acting of graphing, we find that cardinal regulated 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 bespeak of intersection is (3, 2). Thus, (3, 2) should satisfy all equation.
In fact, 3 + 2 = 5 and 3 - 2 = 1
In general, graphical solutions are exclusive approximate. We will break out methods for exact solutions in future sections.
Running equations considered collectively in this fashion are said to physical body a system of equations. A in the above example, the solution of a system of elongate equations can be a single ordered pair. The components of this serial pair satisfy each of the two equations.
Some systems have no solutions, while others undergo an non-finite number of solu- tions. If the graphs of the equations in a system coiffure not intersect-that is, if the lines are parallel (see Figure 8.1a)-the equations are said to be inconsistent, and there is no sequential pair that leave satisfy both equations. If the graphs of the equations are the same line (determine Figure 8.1b), the equations are aforesaid to be dependent, and for each one organized pair which satisfies one equation will satisfy both equations. Mark that when a system is inconsistent, the slopes of the lines are the same only the y-intercepts are different. When a scheme is dependent, the slopes and y-intercepts are the same.
In our work we will beryllium chiefly interested in systems that have one and only one solution and that are said to cost consistent and independent. The graph of so much a system is shown in the solution of Deterrent example 1.
SOLVING SYSTEMS BY ADDITION I
We can resolve systems of equations algebraically. What is to a greater extent, the solutions we obtain by pure mathematics methods are exact.
The system in the following example is the system we considered in Incision 8.1 on page 335.
Example 1
Solve
x + y = 5 (1)
x - y = 1 (2)
Solvent
We can obtain an equation in one variable by adding Equations (1) and (2)
Solving the ensuant equation for x yields
2x = 6, x = 3
We tail now substitute 3 for x in either Equating (1) or Equation (2) to obtain the corresponding value of y. In this case, we have selected Equating (1) and incur
(3) + y = 5
y = 2
Thus, the solution is x = 3, y = 2; or (3, 2).
Notice that we are merely applying the addition property of equality so we can obtain an equivalence containing a single variable. The equation in ace 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 matchless variable by adding Equations (1) and (2) because the price +y and -y are the negatives of each new. Sometimes, information technology is necessary to multiply each member of one of the equations by -1 so that footing in the cookie-cutter variable volition have opposite signs.
Example 2
Solve
2a + b = 4 (3)
a + b = 3 (4)
Solvent
We begin by multiplying each member of Equality (4) by - 1, to obtain
2a + b = 4 (3)
-a - b = - 3 (4')
where +b and -b are negatives of for each one other.
The symbol ', called "prime," indicates an equivalent par; 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 Equality (3) or Equation (4) [say, Equation (4)], we obtain
1 + b = 3
b = 2
and our solution is a = 1, b = 2 Beaver State (1, 2). When the variables are a and b, the sequential pair is given in the variant (a, b).
SOLVING SYSTEMS BY ADDITION 2
As we saw in Segment 8.2, solving a system of equations by addition depends on 1 of the variables in both equations having coefficients that are the negatives of each other. If this is non the case, we can find equivalent equations that do give variables with such coefficients.
Example 1
Solve the system of rules
-5x + 3y = -11
-7x - 2y = -3
Solution
If we multiply each appendage of Equation (1) by 2 and apiece phallus of Equation (2) by 3, we get 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 arrest
-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 price involving variables are in the left-handed member and the constant term is in the right-hand member. We will refer to such arrangements as the standard form for systems. It is expedient 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 criterion 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
-threescore - 2y = -3
and we toilet now proceed as shown above.
SOLVING SYSTEMS BY Permutation
In Sections 8.2 and 8.3, we resolved systems of first-degree equations in two vari- ables by the addition method acting. Another method, known as the substitution method, john as wel make up used to figure out such systems.
Example 1
Figure out the system
-2x + y = 1 (1)
x + 2y = 17 (2)
Root
Solving Equivalence (1) for y in terms of x, we obtain
y = 2x + 1 (1')
We can now ersatz 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
Therefore, the solution of the system is a: x = 3, y = 7; or (3, 7).
In the above representative, it was easy to express y explicitly in terms of x exploitation Equation (1). But we also could have used Equation (2) to write x explicitly in terms of y
x = -2y + 17 (2')
Like a sho substituting - 2y + 17 for x in Equation (1), we begin
Substituting 7 for y in Equation (2'), we have
x = -2(7) + 17 = 3
The solution of the system is again (3, 7).
Short letter that the substitution method is useful if we can easily express one variable in terms of the other variable.
APPLICATIONS USING TWO VARIABLES
If ii variables are related by a single first-degree equation, there are infinitely many consistent pairs that are solutions of the par. But if the two variables are related by two independent first-grade equations, there can be only one ordered geminate that is a solution of both equations. Therefore, to solve problems using two variables, we must constitute two independent relationships using two equations. We can often solve problems more easily by using a system of equations than away exploitation a single equation involving one protean. We will follow the vi steps outlined happening Sri Frederick Handley 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
Stairs 1-2
We make up what we want to find as two countersign phrases. And then, we present the word phrases in terms of two variables.
Smaller number: x
Larger number: y
Step 3 A sketch is not applicable.
Step 4 Like a sho we mustiness compose 2 equations representing the conditions declared.
The sum of two numbers pool is 26.
Step 5 To find the numbers, we solve the organisation
x + y = 26 (1)
y = 2 + 3x (2)
Since Equation (2) shows y explicitly in terms of x, we volition solve the system of rules by the switch method. Substituting 2 + 3x for y in Equivalence (1), we arrest
x + (2 + 3x) = 26
4x = 24
x = 6
Substituting 6 for x in Equation (2), we get
y = 2 + 3(6) = 20
Whole tone 6 The small number is 6 and the larger number is 20.
CHAPTER Sum-up
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Two equations considered together form a system of equations. The solution is generally a single sequential pair. If the graphs of the equations are parallel lines, the equations are same to exist inconsistent; if the graphs are the same personal credit line, the equations are said to be dependent.
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We can solve a system of equations past the addition method if we first-year write the system in classic mould, in which the footing involving the variables are in the left-hand extremity and the constant term is in the right wing-hand member.
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We can solve a organisation of equations past the substitution method if one variable in at least unity equality in the system is first expressed explicitly in terms of the other variable.
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We can solve word problems using two variables by representing two independent relationships by two equations.
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