MARCO TEÓRICO
3.3. El Esqueleto en la Nueva España
Becausea1=0, the first equation has only one variable, and is in the following form:
b1y+c1=0 Solve this equation for y:
b1y= −c1 y= −c1/b1
a2x+b2(−c1/b1)+c2=0 a2x−b2(c1/b1)+c2=0
a2x=b2(c1/b1)−c2 x=[b2(c1/b1)−c2] / a2
2×2 ADDI TION METHOD
Consider the following set of two linear equations in two variables: a1x +b1y+c1=0
a2x +b2y+c2=0
wherea1,a2,b1,b2,c1, and c2are real-number constants, and the variables are represented by xandy. The addition methodof solving these equations consists in performing two separate and independent steps:
•
Multiply one or both equations through by a constant, if necessary, to can- cel out the coefficients of xin the sum of the two equations, and then solve the sum of the two equations for y.•
Multiply one or both equations through by a constant, if necessary, to can- cel out the coefficients of yin the sum of the two equations, and then solve the sum of the two equations for x.To solve for y, begin by multiplying the first equation through by −a2, and the second equation through by a1, and then add the two resulting equations:
(−a2a1x−a2b1y −a2c1=0)
+(a1a2x+a1b2y+a1c2=0) (a1b2−a2b1)y+a1c2−a2c1=0 Then, add a2c1to each side, obtaining:
(a1b2−a2b1)y +a1c2=a2c1 Next, subtract a1c2from each side, obtaining:
(a1b2−a2b1)y =a2c1−a1c2 Finally, divide through by a1b2−a2b1, obtaining:
For this to be valid, the denominator must be nonzero. This requires that a1b2≠ a2b1. If it turns out that a1b2=a2b1, then there are not two distinct solutions to the set of equations.
The process of solving for xis similar. Consider again the original equations: a1x+b1y+c1=0
a2x+b2y+c2=0
Multiply the first equation through by −b2, and the second equation through by b1, and then add the two resulting equations:
(−a1b2x−b1b2y−b2c1=0)
+(a2b1x+b1b2y+b1c2=0) (a2b1−a1b2)x +b1c2−b2c1=0 Then, add b2c1to each side, obtaining:
(a2b1−a1b2)x+b1c2=b2c1 Next, subtract b1c2from each side, obtaining:
(a2b1−a1b2)x=b2c1−b1c2 Finally, divide through by a2b1−a1b2, obtaining:
x =(b2c1−b1c2) / (a2b1−a1b2)
For this to be valid, the denominator must be nonzero. This requires that a1b2≠ a2b1. If it turns out that a1b2=a2b1, then there are not two distinct solutions to the set of equations.
PROBLEM 7-1
Solve the following pair of linear equations (if there is a unique solu- tion) using the substitution method. If no solution exists, or if there are infinitely many solutions, then say so.
3x−6y+9=0
−10x −5y+15=0
SOLUTION 7-1
Let’s begin by solving for xin terms of yin the first equation. Proceed as follows:
3x−6y+9=0 3x −6y= −9
3x=6y−9 x =2y−3
Next, substitute the above-derived solution for x in the second equa- tion, obtaining: −10x −5y+15=0 −10(2y−3)−5y+15=0 −20y+30−5y+15=0 −25y+45=0 −25y= −45 25y=45 y=45/25=9/5
Now that we have the solution for y, we can plug it into either of the original equations and solve for x. Let’s use the second equation. Then:
−10x−5y+15=0 −10x−5(9/5)+15=0 −10x −9 +15=0 −10x +6 =0 −10x= −6 10x=6 x=6/10=3/5 PROBLEM 7-2
Solve the following pair of linear equations (if there is a unique solu- tion) using the addition method. If no solution exists, or if there are infinitely many solutions, then say so.
−4x+y−8=0
−8x +2y−14=0
SOLUTION 7-2
Multiply one of the equations through by a constant, such that one of the variables is canceled out when the two equations are added. Let’s
attempt to cancel out x, getting an equation in yonly. If we multiply the first equation through by −2 and then add it to the second one, we get:
(8x−2y+16=0)
+(−8x+2y −14=0) 2=0
We’ve derived a contradiction. Is there some mistake? Let’s try to can- cel out yinstead, multiplying the second equation through by −1/2 and then adding the two equations. In this case we get:
(−4x +y −8 =0)
+(4x −y+7=0)
−1=0
Again, a contradiction! There is no mistake here. Whenever you use the addition method in an attempt to solve a pair of linear equations and get an equation to the effect that one number equals another, it means that the pair of equations has no solution. Such a pair of equations in two variables is said to be inconsistent.
Occasionally, when applying the addition method to a pair of linear equations, you’ll get another sort of unexpected result: a trivial state- ment that a number equals itself, such as 0 = 0. When this happens, it means that the equations are actually different expressions for the same two-variable equation. In that case, the “pair” of equations has infi- nitely many solutions.
3×3 Linear Equations
A set of three linear equations in three variables presents a more involved (and tedious) problem. The same two general methods, addition and substitution, can be used, but in combination. The following two problems and solutions provide specific examples.
PROBLEM 7-3
Solve the following set of 3×3 linear equations. If no solution exists, or if there are infinitely many solutions, then say so.
2x+5y−z=0
−3x−4y+2z=0 x+y+z=0
SOLUTION 7-3
First, let’s cancel out x, getting an equation in yandz. The coefficients ofxin the above equations are 2, −3, and 1, respectively. When these three coefficients are added, the result is 0. This is convenient, because we can add the three equations together just as they are in order to can- celx. Proceed:
(2x+5y−z=0)
+(−3x−4y+2z=0)
+(x+y+z=0) 2y+2z=0
This result can be manipulated to solve for zin terms of y: 2y+2z=0
y+z=0 z= −y
Now we can reduce the original 3×3 set of equations (in x,y, and z) into a 2×2 set of equations (in xandy). Let’s substitute −yforzin the first two of the original three equations. This gives us:
2x +5y−(−y)=0
−3x−4y −2y=0 which simplifies to this pair of equations:
2x+6y=0
−3x−6y=0
These can be directly added, obtaining the single-variable equation:
−x=0
Multiplying this through by −1 tells us that x =0. That’s one variable down, and two to go! Let’s substitute 0 for x in the first of the above pair of equations in x andy. This gives us:
2x+6y=0 (2×0)+6y =0 6y=0 y=0
Now we can substitute x =0 and y = 0 into any of the original three equations. Let’s use the last one:
x+y+z=0 0+0+z=0 z=0
The solution to this set of 3×3 linear equations, expressed as an ordered triple, is (x,y,z)=(0,0,0).
Did this solution seem “too easy to be true”? In this case, the three equations were selected because there are no constants added or sub- tracted to any of the three equations. That always makes a 3×3 set of linear equations comparatively easy to solve. Now let’s try a more dif- ficult problem.
PROBLEM 7-4
Solve the following set of 3×3 linear equations. If no solution exists, or if there are infinitely many solutions, then say so.
2x+5y −z+2=0
−3x−4y+2z−1=0 x+y+z+5=0
SOLUTION 7-4
First, let’s cancel out x, getting an equation in yandz. The coefficients ofxin the above equations are 2, −3, and 1, respectively. When these three coefficients are added, the result is 0, just as it was in the previ- ous problem. Thus:
(2x+5y−z+2=0)
+(−3x−4y+2z−1=0)
+(x+y+z+5=0) 2y+2z+6=0
This result can be manipulated to solve for zin terms of y: 2y+2z+6=0
2y +2z= −6 y+z= −3
z= −y−3
Now we can reduce the original 3×3 set of equations (in x,y, and z) into a 2×2 set of equations (in xandy). Let’s substitute the quantity (−y−3) forzin the first two of the original three equations. This gives us:
2x+5y−(−y−3)+2=0
−3x−4y+2(−y−3)−1=0 which simplifies to this pair of equations:
2x+6y+5=0
−3x−6y−7=0
These can be directly added, obtaining the single-variable equation:
−x−2=0
−x=2
Multiplying this through by −1 tells us that x = −2. Let’s substitute −2 forxin the first of the above pair of equations in xandy. This gives us:
2x+6y+5=0 (2× −2)+6y+5=0 −4+6y+5=0 6y+1=0 6y= −1 y= −1/6
Now we can substitute x= −2 and y= −1/6 into any of the original three equations. Let’s use the last one:
x+y+z+5 =0
−2−1/6+z+5=0 z−2−1/6+5=0 z+17/6=0
The solution to this set of 3×3 linear equations, expressed as an ordered triple, is:
(x,y,z)=(−2,−1/6,−17/6)